text:
Reusing boiling chips
I was wondering why cannot boiling chips be reused.
As far as I know, they are microporous materials, often coated with PTFE.
They work by, I suppose, giving the heated substance a small cavity, in which a phase transition is favored.
Yet, the golden rule of boiling chips is: they cannot be reused.
I was wondering why they cannot be treated the way dessiccant materials (for instance, molecular sieves) are used, for instance by washing them after use and driying them in oven (especially when non-dangerous materials are involved).

Of course it depends on the use case. Will traces of compounds from the previous use interfere with your requirements for purity in their next usage? Will the chips become coated with some compound from the first usage that interferes with its function as a boiling chip in the following usage? If these things don't apply to your use case, and they appear to be doing their job, then I don't see why they couldn't be used more than once. I have done so on many occasions. Though I don't think I've ever used PTFE coated boiling chips; it seems like that would get in the way of their functionality.

You couldn't coat the chips with PTFE without sealing the pores. That would destroy their functionality.

According to this post at www.reference.com/science

Boiling chips are generally porous, trapping heating bubbles and
  allowing solvent vapors to form in the air cavities. They should
  always be added to a solvent prior to the beginning of a heating
  cycle; adding to a hot liquid may cause it to violently boil over.
  Boiling chips and stones are only used once, as the absorptive
  material becomes impregnated with the first solvent used in.

And Wikipedia confirms the basic idea of why they shouldn't be used more than once:

The structure of a boiling chip traps liquid while in use, meaning
  that they cannot be re-used in laboratory setups.

From personal experience however, I have to disagree with these absolute "cannot be done" sort of statements.  I used to rinse the used chips in an appropriate solvent and then heat them at some temperature well above the solvent's boiling point overnight and don't remember ever having problems.
I realize that the boiling point isn't completely relevant when you can have porous, high surface area material with adsorbed solvent.  Also, as I stated in the comments, I don't remember using PTFE coated chips.

Boiling chips are cheap unless you're using some really exotic type. You'd spend more money on the solvent in which you rinsed them than they are worth. // Having said that, I'd agree that baking them in an oven at a temperature sufficiently above the BP of the solvent should rejuvenate them.


text:
Chile 2017, One of the Largest Fires in South America: Wood Ash Derivates for Industrial Applications
Hi: We're a group of people who wants to help people, domestic animals and wildlife recovering from this fire.

Summary:

We're having in Chile one of the largest fires in South American history.
More than 1,235,527 wood acres have already been burned
The fires aren't stopped but rather increased
People, domestic animals and wild fauna are suffering badly
We're a group that wants to help in the recovery of the lands that have been burned
There're several products with industrial applications that can be extracted from Wood Ashes. For instance, Potassium Peroxide
Our idea is to do the following onsite:

Build over a truck(s) a mobile laboratory
Go to the places where there's too much ashes (making the soil extremely alkaline and hard to recover)
Get some of the ashes, and extract a product with industrial application  that can be sold for a price that produce profitability
The profits would be used for supporting the people who lost everything, helping the wildlife to recover faster and financing the project

The Question:

Are there Wood Ash sub products that you think are suitable to be extracted on site and allow the mentioned objectives mentioned above?
Do you know a technology to be used to do the extraction onsite?
Do you know a chemist who could be interested to participate in the project?

I'm voting to close this question as off-topic because this isn't the type of queries we expect on Chem.SE.

It may be better to do some consulting. Some universities have consulting organizations for physical sciences and mathematics. They may be better equipped to help you. Good luck.

Thanks for your help. I'm an economist and Berkeley MBA. I don't know Chemistry. Fortunately, some Chinese PhDs in Chemistry are already designing a solution! They will bring in the following days a team to help us.  ;)

source: Stack Exchange

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sort: votes


text:
Will fluoride evaporate from tap water at the same rate as chlorine
Chlorine evaporation is an overnight process from standing tap water in a gallon pan. Does fluoride evaporate in the same way/rate?

There's crucial difference between fluorine and fluoride.

Were these downvotes before or after migration? Seems like a good question to me.

And this whole -ine vs. -ide thing is pedantic. Clearly, the asker means the vernacular usages. "Chlorine" and "fluoride" are added to drinking water. How do they compare when using evaporating techniques? I sure hope the downvotes are not because of this.

@fredsbend Chlorination of water is often by adding actual chlorine gas ($\ce{Cl2}$),  whereas fluorination of water can be actual sodium fluoride (NaF) which is a solid.  NaF has no chance of evaporating unless you heat it to over 1000 degrees C.   $\ce{Cl2}$ leaves the water easily like carbon dioxide coming out of soda.

@DavePhD Considering how the question was asked, I think that's exactly the information needed in an answer. The OP is clearly unaware of this. My complaint is if the downvotes are a result of the OP's apparent ignorance regarding what "added chlorine" and "added fluoride" mean chemically.

@fredsbend I upvoted.  It think some people here want to see more college-level chemistry questions maybe, I don't why they downvote really.

Short answer: no, fluoride does not evaporate in the same way/rate as chlorine.
Longer answer: chlorine will be present as covalent, molecular $\ce{Cl2}$. This is, when not in solution, a gas. Chlorine molecules leaving the surface of the water will (on average) not come back (at least unless there is a very high concentration of $\ce{Cl2}$ in the room's air, in which case you have other problems).
Fluoride, on the other hand, is present as $\ce{F-}$ ions (or other, complex ions) along with corresponding positively charged ions (usually $\ce{Na+}$, I believe). Even if the $\ce{NaF}$ were to be removed from solution, $\ce{NaF}$ is a non-volatile salt, and so would not evaporate away.

An imperfect answer because I slightly gloss over how ions and non-ionic molecules behave differently going into / coming out of solution. Suggestions and edits are very welcome!

Considering how the question was asked, I think you answered it perfectly. If the OP wants to know more about the differences between dissolved gases and dissolved salts, then he can look it up or ask a different question.

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text:
Neutralising sulfuric acid but not boric acid
I have another question here which refers to me asking how to separate boric acid from other compounds, and this is partially a follow-up.
Suppose I was to go down the route of making boric acid by the reaction of:
$$\ce{Na2B4O7·10H2O + H2SO4 → Na2SO4 + 4 H3BO3 + 5 H2O}$$
Which in actual fact, I just did 10 minutes ago and failed miserably.
The issue I'm having is that no matter what happens in a home-experiment, you're going to have contaminants, and in this case, they are the sodium sulfate and left-over sulfuric acid.
I added sodium bicarbonate in to remove the sulfuric acid which was stupid because I required so much but then realised that it will also neutralise the boric acid. So next I am going to attempt to neutralise the acid with sodium hydroxide.
But sodium hydroxide will also target the boric acid and just change it back into borax. Is there a way I can neutralise the sulfuric acid without touching the boric acid?

You've got to know that sulfuric acid is a strong acid and boric acid is a very weak acid. So, sulfuric acid can neutralize most of the bases where boric acid cannot. So, you have to take a base that can be neutralized by only sulfuric acid and not boric acid. So, you should probably take group 2 hydroxides and try to react with the unreacted sulfuric acid because sulfuric acid reacts with group 2 hydroxides whereas boric acid can't.
$$\ce{M(OH)2 +  H2SO4 -> M2SO4 + 2H2O}$$
$$\ce{M(OH)2 +  H3BO3 ->nothing}$$
where $\ce{M=Be,Mg,Ca,Sr,Ba}$.
So, which group 2 hydroxide to use? Group 2 hydroxide react with the sulfuric acid to form group 2 sulfates. Now, solubility of group 2 sulfates decreases down the group. So, you should use calcium hydroxide so the calcium sulfate that forms will not dissolve in the solution due to its low solubility and form a precipitate. You can separate that precipitate by filtration.
Formation of calcium borate is highly unlikely and so only calcium sulfate will only form.
Also, barium hydroxide is the best chemical for this process because barium sulfate is very insoluble. But barium hydroxide is costly if you buy it or making it is difficult due to the cost of other reagents. Instead, calcium hydroxide is your best bet.

In actual fact, its a good thing it forms a precipitate, because the water left over in solution will take in the boric acid, and therefore the salt can be filtered off. Should I use maybe calcium hydroxide?

Oh no, I'm just asking weather or not calcium hydroxide will cause something to precipitate out. I don't have any on hand at the moment and since I'm going to be doing a fair few experiments along these lines, I might need to buy or synthesize some.

Solubility of group 2 sulfate decrease down the group. So, if you use calcium hydroxide, calcium sulfate will be formed which is not soluble in water and so, you can separate that precipitate. I will edit the answer.

Great! Thanks alot. Now to go get some calcium hydroxide... :P

Sorry to be a pain, just doing some quick research, and you obviously know much more than me, is boric acid soluble in methanol? Half the MSDS's I find say only ethanol and the other half say yes. :/ Because after filtering out the calcium hydroxide, I'm thinking of just pouring methanol straight in, filtering the other stuff out again, and then just boiling it straight to dry boric acid.

Something's funky here.  Calcium and heavier G2 hydroxides are generally considered strong bases and I would argue that magnesium hydroxide is close to one.  I think they would neutralize the boric acid too.  Beryllium hydroxide might work in principle but beryllium compounds are horrendous from a safety/health point of view.

Sodium bicarbonate is actually a good choice. Bicarbonate ion is a stronger base than both sulfate and bisulfate ions, but a weaker base than borate ion (corresponding to boric acid being only a very weak acid). So the sodium bicarbonate will form sodium sulfate with the sulfuric acid leaving the boric acid unreacted.
Alkaline earth hydroxides, mentioned elsewhere, are not recommended here. Beryllium compounds are toxic and magnesium hydroxide, if dissolved by acid, may be too strong to avoid reaction with the boric acid. Heavier alkaline earth hydroxides would be stronger still.

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text:
What single, pure compound would best replace petrol?
Petrol for use as automotive fuel is produced by blending of different product streams of oil refining. It consists of hundreds of different compounds, and the exact composition can vary widely. Certainly, not all components are absolutely necessary. Is there a single pure compound that could be used instead and that is similar or even superior to typical petrol?
In order to limit the scope of this question, I only consider pure hydrocarbons ($\ce{C_xH_y}$), i.e. no alcohols, ethers, amines, nitro compounds, or other compounds containing hetero atoms. I am only asking about fuel for spark ignition (Otto) engines, i.e. not about diesel fuel.
The relevant criteria for the comparison shall be only the knock resistance and the energy content of the compounds. Other properties may be ignored.
For similar compounds, there may be a tendency for fuel with a higher knock resistance to have a lower energy content. The challenge is to find a single pure compound with a suitable balance of knock resistance and energy content, with better values than found for typical petrol.
The knock resistance shall be expressed as research octane number (RON). Currently, the most important petrol blends in most countries have an RON of 95. However, blends with an RON of up to 102 are widely available. Therefore, the candidate compound should have at least an RON of 102.
The energy content shall be expressed as specific enthalpy of combustion $\Delta_\mathrm ch^\circ$ at a temperature of $T=25\ \mathrm{^\circ C}$. If necessary, this value can be directly calculated from the molar enthalpy of formation $\Delta_\mathrm fH_\mathrm m^\circ$ or the molar enthalpy of combustion $\Delta_\mathrm cH_\mathrm m^\circ$. For this purpose, the generated water shall be assumed to be in liquid form (although this assumption is not representative for vehicle engines and may favour small saturated compounds). The energy content of petrol can vary. For example, typical average values for $\Delta_\mathrm ch$ given by Aral in Germany are

Regular (RON = 91): $-44.2\ \mathrm{MJ/kg}$
Super (RON = 95): $-43.5\ \mathrm{MJ/kg}$
Super Plus (RON = 98): $-42.7\ \mathrm{MJ/kg}$

Therefore, the candidate compound should have at least an energy content corresponding to about $\Delta_\mathrm ch^\circ=-44\ \mathrm{MJ/kg}$.
In order to limit the choice, the candidate compound shall be a liquid at room temperature. The boiling point should be below $210\ \mathrm{^\circ C}$, preferably between $70\ \mathrm{^\circ C}$ and $150\ \mathrm{^\circ C}$. The melting point should be well below $0\ \mathrm{^\circ C}$.
By way of comparison, the following table shows the relevant parameter values for the reference compounds that are used for the octane number (heptane and 2,2,4-trimethylpentane) as well as for one candidate compound (2,2,3-trimethylbutane) that I have already found.
$$
\small\begin{array}{llllllll}
\hline
\text{Compound}&\text{Formula}&\Delta_\mathrm fH_\mathrm m^\circ&\Delta_\mathrm cH_\mathrm m^\circ&\Delta_\mathrm ch^\circ&\mathrm{RON}&T_\mathrm{mp}&T_\mathrm{bp} \\
&&\text{in}\ \mathrm{kJ/mol}&\text{in}\ \mathrm{kJ/mol}&\text{in}\ \mathrm{MJ/kg}&&\text{in}\ \mathrm{^\circ C}&\text{in}\ \mathrm{^\circ C} \\
\hline
\text{Heptane}&\ce{C7H16}&-224.2&-4817&-48.1&\hphantom{00}0&\hphantom{1}{-90.55}&98.4
 \\
\text{2,2,4-Trimethylpentane}&\ce{C8H18}&-259.2&-5461&-47.8&100&-107.3&99.22\\
\text{2,2,3-Trimethylbutane}&\ce{C7H16}&-236.5&-4805&-48.0&112&\hphantom{1}{-24.6}&80.86 \\
\hline
\end{array}$$
I would be interested in answers that search through literature data and find better candidate compounds as well as in answers that use a theoretical approach to estimate where to find the best compounds.

Our city has a some vehicles running on natural gas.  Also if the engines were modified to handle it, hydrogen could be used to fill a cylinder and a spark plug would surely ignite it.

It should also be noted that there is a plethora of other considerations: relative toxicity and carcinogenicity, cost to synthesize, melting/boiling points (for storage), etc.  When looking at a candidate, consumer safety and convenience will also be important.

If this is about fuel surrogates (compositions used to provide a repeatable, specific substitute for a fuel that's variable and tends to degrade) then most of them are mixtures of a small number of hydrocarbons.

Higher $\mathrm{RON}$ seems possible. Yet the boiling point rises accordingly while heat of combustion remains roughly the same. Here are two compounds that fit all posed criteria.
\begin{array}{|c|c|c|c|c|}
\hline
\mathbf{Molecule} & \mathrm{mp\ \mathrm{(^\circ C)}}  & \mathrm{bp\ \mathrm{(^\circ C)}} & \Delta_\mathrm cH_\mathrm m^\circ\ (\mathrm{kJ/mol}) \ \text{@}25\ ^\circ \mathrm{C} & \mathrm{RON} & \mathrm{MON}\\
\hline
\ce{1,3,5-TMB^{[a]}} & -44.7^{[1]} & 164.65(15)^{[1]} & -5193.1(1.3)^{[2]} & 137^{[3]} & 124^{[3]} \\
\ce{2,2,3,3-TMH^{[b]}} & -54^{[4]} & 160^{[5]} & -5405^{[6]} & 112.8^{[3]} & 92.4^{[3]} \\
\hline
\end{array}

[a] 1,3,5-trimethylbenzene, [b] 2,2,3,3-tetramethylhexane

Theoretical predictions of $\mathrm{RON}$
Tareq A. Albahri$^{[7]}$ summarises in his paper the usage of statistics $\mathrm{RON}$ and $\mathrm{MON}$.

The research octane number $\mathrm{RON}$, which is representative of the fuel performance during low-speed city driving, is more often reported in the literature than the motor octane number $\mathrm{MON}$, which is representative of the fuel performance during high-speed highway driving.$^{[7]}$

Theoretical intercalculation between the two characteristics is possible.$^{[7]}$ This work was done by Jenkins$^{[8]}$. First predictions of $\mathrm{RON}$ itself were and are still largely empirical$^{[7]}$. These models use regression techniques to assign effective octane numbers to various organic groups. Albahri examplifies the procedure via Anderson, Sharkey, and Walsh$^{[7]\ [9]}$.

Anderson et al.,$^{[9]}$ for example, developed an empirical model for calculating the RON based on chromatographic analysis of gasoline.$^{[7]}$
In their model, the gasoline is divided into 31 hydrocarbon groups or pseudocomponents, all of which are assigned an “effective” octane number that is estimated by regression of experimental data. The octane number of gasoline is calculated by adding the contribution of octane number[s] from each group.$^{[7]}$

Many similar methods include those proposed by Van Leeuwen et al.$^{[10]}$, Sasano$^{[11]}$, and  Lugo  et  al$^{[12]}$, Ramadhan and Al-Hyali$^{[13]\ [14]}$, Nelson$^{[15]}$, Baird$^{[16]}$,  Twu and Coon$^{[17]\ [18]}$, Rusin et al.$^{[19]}$, Habib$^{[20]}$, Cotterman and Plunkee$^{[21]}.$ Some of these fail to differentiate between isomers, or are valid in specific mixtures.$^{[7]}$ Additionally,

Although  these  techniques [by Van Leeuwen et al$^{[10]}$, Sasano$^{[11]}$, and  Lugo  et  al$^{[12]}$]  give reasonably accurate results, they are usually too time consuming for planning studies and often the compositional data are not available.$^{[7]}$

Tareq A. Albahri's method of SGC (structural group contribution)

After considerable testing, Albahri found that the best equation to take into account contributions of different groups is$^{[7]}$
\begin{split}
  \mathrm{ON} &=
  f\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^{-1}+ a + b\left(\sum_{i}\mathrm{\left(ON\right)_i}\right) \\
  &\quad + c\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^2 + d\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^3 + e\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^4
\end{split}
where $\mathrm{ON}$ is either $\mathrm{RON}$ or $\mathrm{MON}$, and $\left(ON\right)_i$ is the contribution from the $i$th group. Note how the powers increase from left to right.
The coefficients still have experimental basis.$^{[7]}$ This equation is a modified version of what Albahri put forward a year earlier.$^{[22]}$ For many compounds the difference between the estimate and experimatal data is just $5$.$^{[7]}$ If the $\left(ON\right)_i$ are known, one only needs the knowledge of the structure of the molecule to apply Albahri's method.
I suggest reading both papers by Albahri since they include tables of data and an example calculation.
Here is a selection of $\left(ON\right)_i$ values.$^{[22]}$

QSPR Model for octane number prediction

In a research article by Al-Fahemi, Albis, Gad, a quantitative structure-property relationship (QSPR) is performed. This includes establishing a correlation relation between $\mathrm{ON}$ and various physical parameters.$^{[23]}$ Again the model is based upon regression rather than derivation from first principles.
\begin{align}
\mathrm{ON} =
    & - (193.53 \pm 319.19) + (1.47 \pm 1.01)M\\
    & - (53.06 \pm 31.47)E_H - (8.67 \pm 2.73)B_P\\
    & - (24.94 \pm 19.44)M_R - (50.52 \pm 26.92)\log P\\
    & + (4.33 \pm 3.09)C_P + (3.72 \pm 2.04)C_V + (5.17 \pm 2.08)C_T\\
\end{align}
where

$M$ is molecular mass,
$E_H$ is hydration energy,
$B_P$ is boing point,
$\log P$ $-$ octanol/water distribution coefficient,
$M_R$ is molar refractivity,
$C_P$, $C_V$, $C_T$ is the critical point (pressure, volume and temperature, respectively).$^{[23]}$

TL; DR
There is no model to date that estimates or derives a result for $\mathrm{ON}$ from first principles. Different models depend more or less on multiparameter regression analysis and experimental data. Nevertheless, the methods of Albahri and others have predictive capability. Often the difference beteen estimated and experimental octane numbers is less than $5$.
Or as Albahri mentions,

Octane number is one of the most difficult properties to estimate or correlate because of its complex dependency on the molecular structure of the compound.$^{[7]}$
An estimation technique of the octane rating of pure hydrocarbons, though  essential, is nonexistent.$^{[7]}$

So for your second question, I suggest applying Albahri's method because

widest applicability among hydrocarbons,
relatively easy to use,
few $\mathrm{RON}$s are known from experiments,
Al-Fahemi et al.'s procedure requires many parameters which will be a pain to track down (if known at all).

Qualitative considerations$^{[3]\ [7]}$

shorter alkane chain $\ce{->}$ higher $\mathrm{ON}$ (but higher volatility)
branching of alkane $\ce{->}$ higher $\mathrm{ON}$
aromaticity in hydrocarbon $\ce{->}$ higher $\mathrm{ON}$ (but tends to burn sooty with carcinogenic byproducts)
effects on $\mathrm{RON}$ and $\mathrm{MON}$ can be very different

Arithmetic average of $\mathrm{RON}$ and $\mathrm{MON}$ $-$ anti-knock index or $\mathrm{AKI}$ $-$ is also used.
$$\mathrm{AKI} = \frac{\mathrm{RON} + \mathrm{MON}}{2}$$

$[1]$
T.F. Ardyukova, I.K. Korobeinicheva, A.I. Rezvukhin. Atlas of
Spectra of Aromatic and Heterocyclic Substances. $(1973)$, 7.

$[2]$
W. H. Johnson, E.J. Prosen, F.D. Rossini. 'Heats of combustion and
isomerization of the eight $\ce{C9H12}$ alkylbenzenes', Journal of
Research of NIST, $(1945)$, 35, pp 141$-$146.
$[3]$ A.
Demirbas, M. A. Balubaid, A. M. Basahel, W. Ahmad, M. H. Sheikh.
'Octane Rating of Gasoline and Octane Booster Additives'. Petroleum
Science and Technology, $(2015)$, 33(11), pp 1190$-$1197. DOI:
10.1080/10916466.2015.1050506.

$[4]$
Jean-Claude Bradley Open Melting Point Dataset. (January 29, 2017)

$[5]$
R. Stenutz. 2,2,3,3-tetramethylhexane. A collection of tables for
chemistry. (January 29, 2017)

$[6]$
Pure Components Data. (Excel file.) Petroleum Engineering, Texas A&M
University. (January 29, 2017)
$[7]$ Tareq A. Albahri. 'Structural Group Contribution Method for Predicting the Octane Number of Pure Hydrocarbon Liquids'. Industrial & Engineering Chemistry Research $2003$, 43(24). DOI: 10.1021/ie020306+.
 $[8]$ G. I. Jenkins. 'Calculation of the motor octane number
from the research octane number'. Journal of the Institute of
Petroleum, $1968$, 54, 14.

$[9]$
Anderson, P. C.; Sharkey, J. M.; Walsh, R. P. 'Calculation of  the
Research  Octane  Number  of  Motor  Gasoline  from  Gas
Chromatographic Data and a New Approach to Motor Gasoline Control'.
Journal of the Institute of Petroleum, $1972$, 58, 83.

$[10]$
J. A. Van Leeuwen, R. J. Jonker, R. Gill. 'Octane Number Prediction
Based on Gas Chromatographic Analysis and Nonlinear Regression
Techniques'. Chemometrics and Intelligent Laboratory Systems,
$1994$, 25(2), pp 325$-$340. DOI: 10.1016/0169-7439(94)85051-8.

 $[11]$  Y. Sasano. 'Measuring Research Octane Number of
Gasoline by Gas Chromatograph'. JP Patent 09138613, $1997$.
 $[12]$ H. J. Lugo, G. Ragone, J. Zambrano. 'Correlations
between Octane Numbers and Catalytic Cracking Naphtha Composition'.
Industrial & Engineering Chemistry Research, $1999$, 38, 2171.
 $[13]$ O. M. Ramadhan, E. A. S. Al-Hyali. 'New Experimental and
Theoretical Relation to Determine the Research Octane Number (RON) of
Authentic Aromatic Hydrocarbons that Could be Present in the Gasoline
Fraction'. Petroleum Science and Technology, $1999$, 17, 623.

 $[14]$ O. M. Ramadhan, E. A. S. Al-Hyali. 'Aromatic
Hydrocarbons in Some Iraqi Gasoline and Their Influence in the Value
of the Research Octane Number'. Petroleum Science and Technology,
$1999$, 17, 607.
 $[15]$ W. L. Nelson. 'Octane Numbers of Naphthas'. Oil and Gas
Journal, $1969$, 67, 122.
 $[16]$ C. T. Ch Baird. Address: De  la  Haute-Belotte  6,  1222
Vezenaz, Geneva, Switzerland. Unpublished results.
 $[17]$ C. H. Twu, J. E. Coon. 'Predict Octane Numbers Using a
Generalized Interaction Method'. Hydrocarbon Processing, $1996$, 71,

$[18]$
C. H. Twu, J. E. Coon. 'Estimate Octane Numbers Using an Enhanced
Method'. Hydrocarbon Processing, $1997$, 76, 65.
 $[19]$ M. H. Rusin, H. S. Chung, J. F. Marshall. 'A
Transformation  Method  for  Calculating  the  Research  and  Motor
Octane Numbers of Gasoline Blends'. Industrial and Engineering
Chemistry Research. $1981$, 20, 195.
 $[20]$ E. T. Habib. 'Effect of Catalyst, Feedstock, and
Operating Conditions  on  the  Composition  and  Octane  Number  of
FCC Gasoline'.  ACS  Symposium  Division  of  Petroleum  Chemistry,
Miami, FL, Sept $1989$.
 $[21]$ R. L. Cotterman, K. W. Plunkee. 'Effects  of  Gasoline
Composition  on  Octane  Number'.  ACS  Symposium  Division  of
Petroleum Chemistry, Miami, FL, Sept $1989$.
$[22]$ Tareq A. Albahri. 'Structural Group Contribution Method for Predicting the Octane Number of Pure Hydrocarbons and Their Mixtures'. Fuel Chemistry Division Preprints, $2002$, 47(2), 532
$[23]$ Jabir H.
Al-Fahemi, Nahla A. Albis, Elshafie A. M. Gad. 'QSPR Models for
Octane Number Prediction'. Journal of Theoretical Chemistry,
$2014$, vol 2014, article ID: 520652, 6 pages. DOI: 10.1155/2014/520652.

Even longer branched hydrocarbon chains or benzene derivatives will probably have higher $\mathrm{RON}$s. Finding data gets harder by each carbon atom though.

To reiterate, I think Tareq A. Albahri's method is the best we have got so far.


text:
How to create 1-100 ppm ethanol gas?
I am trying to use TGS 2602 sensor of air contaminants. I would like to test if my circuit is working as intended. For that purpose I need to generate fumes of one of the gases the can be detected by the sensor.

I thought that I would go for ethanol. I tried holding vodka soaked cotton ball near the sensor - to no avail. Later on I tried spilling vodka around the circuit and placing more vodka soaked cotton balls around. I also covered the whole thing with a cardboard box (with gaps at the bottom between table and the box). Again there was no considerable effect. Vodka is essentially ~40% ethanol and water mix (by volume).
Can you say by that short description that I managed to generate between 1 and 100 ppm ethanol gas? If not, can you propose a robust and safe method of producing such a gas?
Please be aware how the sensor looks. It is enclosed in a metallic case with small holes on the top surface.

Are you certain that the wiring and power supply 5 V DC for sensing and for heating are correct? You might want to look at the datasheet.

@Klaus Warzecha - that is what I am trying to establish. I am using a somewhat more complicated circuit, because I am using digital rather than analog reader. Anyway when I replace the sensor with resistors (~57 ohm for the heater and ~10k or ~100k ohm for the sensor itself) the circuit seems to work - nevertheless I am not 100% sure that it is functioning as intended, therefore I would like to create a testing environment in which I should have prior knowledge that sensor should be sensing something if it was working correctly.

How to do it, and how to do it cheap are two very different problems.

If the 40% alcohol solution came to vapor-liquid equilibrium at room temperature , then the vapor concentration of ethanol should have been on the order of 2%.

Waving a vodka soaked cotton ball near the sensor is sufficient to generate the required fumes. It was the circuit that was problematic - I did a subsequent testing with analog reader (voltmeter to be specific).


text:
Why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?
On the Pauling electronegativity scale, fluorine and chlorine are 3.98 and 3.16, respectively.
Since the dipole moment is dependent on electronegativity, why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?
I guess that the smaller size of the s orbital in the fluorine atom has something to do with this.

Dipole moment is not just about charges, it also has $L$ term. Bond length of $\ce{C-Cl}$ is greater than $\ce{C-F}$ and in this case, that is more dominating factor.
The dipole moment is in order
$$\ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I}$$
You can see that electronegativity plays a more dominating role in $\ce{CH3X}$ when $\ce{X}$ is $\ce{Br}$/$\ce{I}$.


text:
Why is temperature considered as an important factor in deciding chemical reaction kinetics?
According to the Arrhenius equation, the rate of reaction is proportional to $e^{-1/RT}$.
Here is a plot of the equation and its derivative with respect to temperature, both with respect to temperature. All constants are set to unity.

It seems to suggest that there is almost no change in the rate of reaction with increasing temperature except at very-low-temperature range.
Then, why is temperature considered an important factor in deciding the kinetics of a chemical reaction?

The Arrhenius eqn. describes how the rate constant changes with temperature as $k=A\exp(-E_a/(RT))$ where $E_a$ is the activation energy (an energy barrier) and A is a constant. At higher temperatures more molecules (by the Boltzmann distribution) have enough energy to surmount the activation barrier and so reaction becomes faster. Negative temperature as in your figure do not make sense. At very high temperatures the exponential tends to 1 as all  molecules have enough energy to exceed the barrier energy and so the rate constant tends to A.

It is possible to produce negative temperatures, (inverted population) such as in a laser, in certain NMR experiments, and in a chemical laser. But the first two are not chemical reactions and the latter the population inversion is in the products of an exothermic reaction. To answer your last question the rate constant does change by a large amount from low to high temperatures, basically from zero to from finite limiting value at high temperature.

If the activation energy is lower (<<50kJ/mol),  then temperature has a much lower effect than when activation energy is higher (>>100kJ/mol). Your comment would be true for the former case


text:
Reasoning behind IUPAC nomenclature?

What is the reason for writing prefixes like di and tri in front of prefixes of alkyl or functional groups ?

For example in 2,3-Dimethylpentane we write 2,3 which is the location of the methyl group in the parent chain but then we also write Di to indicate the presence of two methyl groups.
I do not understand why we write Di, isn't it obvious that if there are two locants given then there will be two methyl group ? one methyl group is not going to be present at two places simultaneously.

Ok now lets assume for a second that writing Di is somehow useful, then
in 2,2-Dimethylpentane why we write 2,2 ? for me it is obvious that if there is Di before the methyl and only one locant is given then two methyl groups will be present at that location.

To summarise :

In 2,3-Dimethylpentane why we write Di if we have already written two locants to indicate the location of two methyl groups ?
In 2,2-Dimethylpentane if we write Di then why we write 2,2 ? I think only 2 is suffice here.

Related: Why do we insert a numeric multiplier after group positions when naming an isomer?

@Loong I got the answer for my first question but not for the second question.

This is of course conjecture, but a point would be to make it easier to "say" the compounds. The system would be weakened if, by intuition, people called substances by different nomenclature than what they wrote. If iupac nomenclature forced us to call something methyl-methyl-pentane we would simply say dimethylpentane anyway. Might as well accomodate easy spoken names and include the prefix as well as the numerical. This way there is 1 system for all, no exceptions. Fewer exceptions means stronger system.

@StianYttervik That might explain the first part but what do you think of second part ?

I mean that it is easier for everyone if we just keep both the "oral" tradition (di-, tri-) and the numeric ones as the standard. If there were to be exceptions when it didn't mean anything it would be confusing and had to be decided on a case for case basis whether or not to use the numerics. Better to just always have them, they usually represent an insignificant waste of bytes...


text:
Which metals hydroxides dissolve in excess NH4OH?
Some metal hydroxides like $\ce{Ag(OH)}$ and $\ce{Cu(OH)2}$ dissolve in excess $\ce{NH4OH}$ to form metal ammonia complexes.
I want to know which other common metal hydroxides dissolve in excess $\ce{NH4OH}$. Also, can it be somehow predicted ? I couldn't find much relevant information on the Internet.

There are a number of unknowns and a number of equations but this is a matter of the various equilibriums. Remember that you could also add NaOH so the solution could be more basic than you could get with NH4OH alone.

The Wikipedia article for "Metal amine complex" gives many examples.  According to the article:

Almost all metal ions bind ammonia as a ligand, but the most prevalent
  examples of ammine complexes are for Cr(III), Co(III), Ni(II), Cu(II)
  as well as several platinum group metals.

Specific examples include:
$\ce{[Pt(NH3)4][PtCl4]}$
$\ce{[RhCl(NH3)5]Cl2}$
$\ce{[Cr(NH3)4Cl2]Cl}$
$\ce{[Co(NH3)6]Cl3}$
$\ce{NH4[Cr(NCS)4(NH3)2]}$
$\ce{[Zn(NH3)4]^2+}$
$\ce{[Ag(NH3)2]+}$
$\ce{[Ni(NH3)6]^2+}$
$\ce{[Cu(NH3)4(H2O)2]^2+}$
Several others are listed just in that one Wikipedia article, along with a fair list of references.


text:
Why aren't these two resonance structure same?

Draw the resonance structures of Fluorobenzene.

In the solutions the structures below were given, I wonder why these structures are not just same. Both of these structures have a negative formal charge on the ortho position.

Are the above structure of Fluorobenzene same ?

They are not "really" different, but there are two ortho carbons in fluorobenzene so you draw a resonance structure showing charge delocalization to each ortho carbon.  There is also a third resonance structure with the charge on the para carbon - 3 resonance structures in all.

@ron Does that mean that when we draw resonance structures we aim to show all the different places where electron can delocalize ? does not matter if it is resonance structures are similar or different.

Yes, that's correct.

@ron Thank you $\ddot \smile$. That instantly solved many of my questions about resonance theory.

Simply - When you're drawing resonance structures you effectively number the atoms to break symmetry.

There are called equivalent resonance structures since the number of pi bonds, lone pairs and sigma bonds are the same. On the other hand, resonance is the delocalization of pi electrons. These structures show how pi electrons are delocalised from ortho position to para position and then back to ortho(on the other side of the molecule). So both the structures you mentioned are different in this context and do have a purpose in explaining the resonance phenomenon.0
Note: This rule only applies to hydrocarbons. In other cases, the two resonance structures need to have the same structural formula to be equivalent.

"There are called equivalent resonance structures since the number of pi bonds, lone pairs and sigma bonds are the same. " Does not resonance structures ought to have number of pi, sigma and lone pair equal ? I maybe wrong.

Two resonance structures must obey this condition: they must have the same number of electrons involved in sigma bonds, pi bonds and lone pairs. If they also happen to have the same number of pi bonds, lone pairs and sigma bonds, they are called equivalent resonance structures. Every resonance structure need not be a equivalent resonance structures but every equivalent resonance structures must be a resonance structure.

What about these? Are they equivalent? They have the same number of pi bonds, lone pairs and sigma bonds, but they are most certainly not equivalent.

@orthocresol Oh yes I think you are correct. So what is the criteria for equivalence ?

That both resonance forms should have the same structural formula.


text:
derivation of partial differential equations for changes in U

Changes in internal energy at

At constant temperature, $$d U = \left({\partial U \over \partial V}\right)_TdV$$
At constant volume,
  $$d U = \left({\partial U \over \partial T}\right)_VdT$$
When both temperature and volume is changing,
  $$d U = \left({\partial U \over \partial V}\right)_TdV +\left({\partial U \over \partial T}\right)_VdT$$

I want to know how do we get these equations.

I think I can speculate a bit about how do we get the first and second equation.
For the first equation, since the temperature is kept constant, we can plot a U-V graph.
Tell me if I am wrong here but for that graph the slope at a given point is $\displaystyle \left({\partial U \over \partial V}\right)_T$.
Now since the change is a infinitesimal change, the slope of tangent at that point is given by $\displaystyle \left({\partial U \over \partial V}\right)_T$.
So $\displaystyle {\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$, since the change is infinitesimal, $\Delta U = dU, \Delta V = dV$ and substituting this we will get out equation 1. Same can be said for equation 2.
I have no idea how do we get the equation 3, I think it is a clever use of  product rule or something like that.

Is my understanding for the equations 1 and 2 correct ?
How do we get equation 3 from 1 and 2 ? because simply adding 1 and 2 does only give the RHS of 3 not the LHS.

One definition of a state function such as internal energy $U$ is that its infinitesimal change is equal to an exact differential. Thus in general
$$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$
There are various variables with regard to which one can apply the partial derivative. Entropy, volume, temperature to name a few. Usually we assume that our thermodynamic state functions are two-dimensional but this is only if the number of moles is constant.
$$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i \overset{\mathrm{d}n_j\ =\ 0}{\implies} \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V.$$
A good way to think of this assumption is that other functions are also simply functions of $T$ and $V$, e.g., $S = S(T,V)$. From the equation
$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V$$
you can reach the other two equations by imposing $\mathrm{d}T = 0$ and $\mathrm{d}V = 0$, respectively.

So my way of thinking of how first and second equation is incorrect ?

Also can please you explain a bit more about how did you get the second equation. I did not understand that.

@A---B You are right that if temperature is constant, then you are essentially in a sliver where $U$ varies as $V$ is changed. And yes, the derivative $\left({\partial U \over \partial V}\right)_T$ will thus be the "slope" in that particular direction. Using $${\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$$ is not a rigorous way of thinking about it though. I would focus on the general case (the total or exact differential) and derive equations $1$ and $2$ from there.

Mostly likely you are correct. But I did not understand from where did you get  $$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i$$ and what does $x_i$ mean here ?

@A---B I assume that $U = U(T, V, n_j)$ and calculate a total differential. My point was that there is no need to add partial derivatives with respect to entroy etc. because $S = S(T, V, n_j)$ is already included in the $k + 2$ partial derivatives.

@A---B It has $n_j$-s, (number of moles) it is indeed smally written and I was afraid that it is not easy to see :)

I did not see that. I was thinking about $x_i$ in $$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$. What is x here ?

@A---B Oh, just a random variable. Later on I took $x_1 = T, x_2 = V, x_3 = n_1$ and so on.

@A---B Glad I could help!


text:
What is the physical interpretation of "internal pressure"?

$$ \require{begingroup}
\begingroup
\newcommand{\md}[0]{\mathrm{d}} \md U = \pi_T\,\md V + C_V \,\md T$$
  where  $\pi_T$ is the internal pressure and is given by $\displaystyle {\left(\partial U \over \partial V\right)_T}$.

What does the internal pressure actually physically mean? I can see it is the slope of the $U$-$V$ graph at constant temperature but that does not clear things up.
Also why is this quantity called the internal pressure and not something else? $\endgroup$

I'd assume that the phrase is trying to acknowledge the wall effect. Pressure at the wall of the container isn't the same as pressure away from the wall.

@MaxW Outside the container or inside the conatiner ?

Inside of course. what would the pressure outside of the container have to do with anything?

I also thought that but I did not understand where that pressure will be acting ?  on other molecules ?

I don't understand your last remark. Could you please clarify what "that pressure" is?

@MaxW I am really sorry. The pressure that is away from the wall.

Let's look at it this way. I am on the starship Enterprise and contain a 1 meter sphere of gas in a force field. I then beam the sphere into deep space and then release the force field. What happens? The gas dissipates in a spherical direction since there is nothing pushing the gas back together.

@MaxW So that force field is internal pressure ?

No, the force field acts against the internal pressure.

@MaxW Yes sorry, my mistake I thought the force field is inside the sphere of gas, whereas it is the exact opposite.

See orthocresol's derivation of an internal pressure $\pi_T$ relation $(1)$ here.
$$\pi_T = T\left(\frac{\partial P}{\partial T}\right)_V - P\tag{1}$$

Internal pressure $\pi_T$ measures what you would expect from its definition:
$$\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T.$$
It shows how internal energy changes when volume is changed and temperature is constant. For ideal gases the change is zero, something that also follows from the equipartition theorem.
$$\mathrm{d}U_{\text{ideal gas}}= \frac{\nu}{2}nR\mathrm{d}T \implies \left(\frac{\partial U}{\partial V}\right)_T = 0 \ \  \text{for an ideal gas}$$
In other words, the total energy an ideal gas has is not dependent on volume. There is no repulsion nor attraction amongst ideal gas particles.
As a first approximation to model real gases, the van der Waals equation $(2)$ is used.
$$\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\tag{2}$$
Hence,
$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$
Using relations $(1)$ and $(3)$
$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V-nb} \overset{(3)}{=} \frac{1}{T}\left(P + a\frac{n^2}{V^2}\right).$$
Again via formula $(1)$
$$\pi_T = T \cdot \overbrace{\frac{1}{T}\left(P + a\frac{n^2}{V^2}\right)}^{(\partial P/\partial T)_V} - P = a\frac{n^2}{V^2}\tag{4}.$$
Result $(4)$ $\pi_T = a\frac{n^2}{V^2}$ demonstrates that for real gases internal energy will change when volume is decreased or increased. Real particles have interparticular forces that vary with distance. A hint as to why $\pi_T$ is called internal pressure will be given when $(4)$ is substituted into $(2)$:
$$\left(P+\pi_T\right)\left(V-nb\right)=nRT,$$
or even better, $(3)$:
$$P = \frac{nRT}{V-nb} - \pi_T.$$

If $\pi_T > 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
If $\pi_T < 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.

I don't think this is what the OP is asking about. I think question is more about why "internal pressure" vs. just "pressure." In other words what does "internal" add?

@MaxW I did always think about it as 'a correction to pressure that derives from intermolecular forces'. As opposed to traditional (ideal gas) pressure which can be modelled as collisions with container walls. To recap, real gases occupy a finite volume, and if one decreases the volume of the system (via an external pressure), the molecules are forced more close together. How these particles respond is what internal pressure measures. This is, in turn, to be differentiated from the traditional pressure (of which internal pressure is a component).

@MaxW Whether this is a good way of thinking about it, remains to be seen. :=) (Feel free to downvote accordingly if this is wrong.)

@LinearChristmas Wow, this is easily one of the most complete answers I have ever got. Crystal clear. Thanks for the answer.

That is pretty much the significance of $\pi_T$, it is a measure of intermolecular (IM) forces. As V changes, the average IM distance will change, and then the difference in the IM potential energy will affect U. For an ideal gas, there are no IM forces; hence there's no effect on U and $\pi_T = 0$. (And I realise I am just repeating what you said... whoops)

Assume that the gas is enclosed in a cylinder with a movable piston but the piston is not moving now. If the piston - wall interface is frictionless then the internal pressure (the one you are asking about) is the same as the external pressure (the one that can be controlled by mechanical means, e.g., by pushing or pulling on it.) If there is no friction and the process is reversible then the internal pressure is equal to the eternal pressure (the control variable) and is a state variable one that can be described by some other equilibrium parameters, such as $U$ or $T$ or $V$. If there is friction between the piston and the wall then depending on whether one pushes the piston in or pulls it out the internal pressure maybe larger or smaller than the external one and there is no such functional relationship among these parameters.


text:
Work done under constant pressure

Work done under a constant pressure.

I know $$\Delta U = \Delta V \times \pi_T + C_V\Delta T $$
I thought I can give a try to find work done from above equation under constant pressure for an ideal gas.
$$q_p + w_p = C_V\Delta T$$
$$\implies C_p\Delta T + w_p = C_V \Delta T$$
$$\implies w_p = -\Delta T (C_p - C_V)$$
$$\implies w_p = - R\Delta T $$
Adiabatic work is $w_\mathrm{ad} = C_V \Delta T$
For a adiabatic work under constant pressure $w_\mathrm{ad} = w_{q} \implies C_V = -R \implies C_p = 2R$
Constant $C_p, C_V$ and negative $\gamma$ does not look correct to me.

Which of my steps are incorrect ?

You can't do adiabatic reversible work at constant pressure.  The pressure has to change.  If you want to do adiabatic irreversible work at constant pressure, say by suddenly increasing or decreasing the pressure load to begin with, and then holding it constant during the process, this is sometimes referred to as adiabatic work at constant pressure, but it really isn't.  The final pressure will not be equal to the initial pressure that was present before you changed the applied load.

So I should use Kirchoff's law to determine $q_p$ ?

I also did not understand why work can't be done on a adiabatic system under constant pressure.

In a closed system, the only way to compress or expand a gas such that the initial pressure is equal to the final pressure, and is constant along the entire path is to raise its temperature by adding heat.  But that's not adiabatic.  Try solving a reversible adiabatic path, and see how that plays out.

Ok I will try that out. Is my statement, $w_p = - R\Delta T$, correct for work done under constant pressure for an ideal gas ?

Yes, if that's the work done per mole by the surroundings on the gas.  Don't forget that, in an adiabatic reversible process, the pressure is changing, so you have to take the integral of PdV=RTdV/V to get the work.

Thank you, I will tell you what I got. I find $w_p = - R\Delta T$ fascinating , It is pretty amazing.

That's only if the pressure is truly constant throughout the process.

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


text:
Why is E2 reactivity rate of 1-iodo-1-methylcyclohexane faster than (1R,2S)-1-iodo-2-methylcyclohexane even though the transition state is same?

Why is E2 reactivity rate of the second compound said to be faster than the first compound even though the transition state is same? My teacher said that in E2 more hindrance implies faster reaction. In the first compound H atom is more hindered due to methyl group. Even then why is the reactivity of the second compound in presence of a strong base, said to be faster?

See: E1 and E2 elimination reactions of isotopically labeled bromocyclohexanes

If the rates are different, the transition state must be different, no?

Why do you think the transition states are identical? The connectivity is different...


text:
Oxidation state of Halogens
So the question is which is the most common oxidation state of halogens and I have to justify my answer. Is my following thought correct? Do I have to add something?
The most common oxidation state of halogens is $-1$. Halogens are the most electronegative elements of the periodic table. Their outer electron configuration is $n\mathrm{s^2}n\mathrm{p^5}$. If chlorine, for example, gains one more electron, the outer p orbitals are completely filled (resulting in a full octet). Since halogens are electronegative they can easily remove an electron from the nearby atom. So then the halogen becomes isoelectronic with a noble gas and it is more stable in this state.

Whatever you have thought is correct.

@Prakhar No, that is wrong. Please do not confuse people by posting wrong comments.

Well ,we are talking about the most common oxidation state of Halogens, and -1 is the most common oxidation state. I know that they show oxidation states from -1 to +7 (except for fluorine), but that is not what we are being asked here. We are being asked about the oxidation state that Halogens exhibit more frequently or the oxidation state that is observed mostly.

@Prakhar Again, this is wrong. Iodide is not iodine’s most common oxidation state.

I suspect the "book" answer is -1, but that does not make it really right for all halogens especially iodine and heavier.  "Dumbing down" science for introductory courses is something we have to put up with :-( .

One curious case is only a single halogen is known in a compound with zero oxidation state. That would be fluorine in "gold heptafluoride", $\ce{AuF5•F2}$. Everything else with the halogens is known only as $-1$ or positive, and predominantly odd.

There is not one single most common oxidation state across all halogens. While fluorine is the most electronegative element of which compounds are known — meaning that it can only exist in the $\mathrm{-I}$ and $\pm0$ oxidation states — the electronegativity continually reduces down the group to iodine. Iodide in the oxidation state $\mathrm{-I}$ is easily oxidised to $\ce{I2}$ or $\ce{IO3-}$ with the oxidation states $\pm0$ and $\mathrm{+V}$, respectively.
What is true, however, is that all halogens except fluorine show oxidation states from $\mathrm{-I}$ to $\mathrm{+VII}$ as can be predicted from their electronic structure $n\mathrm{s^2}\,n\mathrm{p^5}$.

Hmm, plus/minus zero?

@Mithoron Yes …?

sigh That meant that it has no sense whatsoever in chemistry to put such sign in front of zero (sort of euphemism ;)

@Mithoron I actually learnt to always label the oxidation state as $\pm0$. Because all other oxidation states are signed, that one should be, too. And also, all other oxidation states are written in Roman numerals but there is no Roman numeral zero — having something that stands out makes it clearer we are dealing with an oxidation state. All of my answers that talk about an oxidation state $\pm0$ should have that sign; if not it is a mistake on my behalf and I shall correct it asap ;)

In the wikipedia article, it's written $0$ everywhere. @Jan


text:
VSEPR geometry of Bromate Ion
After doing the Lewis structure of Bromate Ion and in order to find the molecular geometry using VESPR method, we have:
central atom: Br
Electrons of the central atom: 7
Electrons that contribute the 3 Oxygens : 3
Electrons that contribute the central atom for the π bonds: -2
Charge of ion (negative) : 1
Total electrons : 9
But if we divide 9 with 2 in order to find the σ-bonding pairs, we get 4,5. what is wrong? Because after searching the molecular geometry of BrO3- it says that it is trigonal pyramid.

I'm not sure what is your difficulty exactly. If you've already done the Lewis structure, then you know the Bromate ion looks like this.  Three terminal atoms + a lone pair indicate that this polyatomic ion is of the type AX3E1. According to this table, this is a trigonal pyramid. So what exactly is the problem? What purpose do these calculations serve?

@Jan I am a bit hesitant to agree that this is a duplicate. The reason for this is that the target question is convoluted with needless information of a pupil-teacher e-mail back and forth and it is hard to find the actual question. I like your answer there and would prefer to see it here.

@Martin-マーチン Fairy ’nuff ;)

There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:

How many valence electrons do we have? (add up)
For bromate, the calculation is:
$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$
How many valence electrons are required for full octets/hydrogen dublets?
This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.
$$4\times 8 = 32\tag{2}$$
How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$
$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$
How many electrons are remaining? These will form lone pairs. $(1) - (3)$
$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$

Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{Br-O}$ bonds are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure:

(Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image. Originally taken from an answer about iodate hence the central iodine. Replace it with bromine to get bromate.)

Thank you, that was very helpful! According to this, the molecular geometry of ClO2- will be linear? valence electrons=20, electrons we need to form octets: 3x8=24 , electrons we're missing : 24-20=4 or 2 pairs and electrons that will form lone pairs: 20-4=16 or 8 pairs

@MariaP No, it will be bent; you will have two lone pairs on chlorine (and three per oxygen). Linear compounds are very rare for purely single-bonded atoms; to the best of my knowledge, they only occur with beryllium. Multiple bonds are a different story.


text:
Why is the hydbridization of the N atom (rightmost) in diazomethane sp and not sp2?
Why is the hydbridization of the $N$ atom (rightmost) in diazomethane $sp$ and not $sp^2$ ?
Most books claim that the negative charge (electron pair) of nitrogen can participate in resonance and is placed in a pure $p$ orbital. However, according to the general rule the hydribization number (number of sigma bonds + number of lone pairs) of the $N$ comes out to be $3$ implying ($sp^2$). However, according to several books the hybrdization is $sp$ and not $sp^2$. Why is it so?

As always in these questions, draw the resonance structure.

@Zhe I don't agree with you. We can draw a resonance structure only if we know beforehand that the electron pair is in pure p orbital. But, unless we know that the hybridization is $sp$ we cannot do so. If both the electron pairs were in hybrid sp2 orbitals then no resonance would have taken place.

You don't have to agree with me, but it does. That's why diazomethane behaves like a carbene and is nucleophilic at the carbon.
In addition, the logic here is backwards. You should be thinking about resonance and whether the electron should go into a $p$ orbital to begin with, not the other way around.


text:
Why do we use the external pressure to calculate the work done by gas
I read in a textbook that in the case when we have a gas in a cylinder fitted with a massless frictionless piston being held with an external pressure $p_1$, and when the pressure is reduced to become the value $p_2$, the gas pushes up against the piston and then the work done by the gas for a small change in volume is calculated by:
$$\mathrm dW=p_2\,\mathrm dV$$
Here, is what I don't conceptually understand. If the gas's molecules was under some pressure $p_1$ which is equal to the external pressure in the static state then after the external pressure became lower than the internal pressure, shouldn't the work done by the gas be the difference between the two pressures?

If the piston is frictonless and massless, then, if you do a force balance on the piston, you must have that the force per unit area that the gas exerts on the inside face of the piston will always be equal to the external force per unit area that one imposes on the outside face of the piston.  The sudden drop in pressure on the outside face of the piston causes the gas to undergo an irreversible expansion.  During an irreversible expansion, the local pressure within the cylinder becomes non-uniform, so that the average pressure of the gas differs from the force per unit area at the piston face.  As a result, the ideal gas law (or other equation of state) cannot be applied globally to the gas in the cylinder.  In addition, during an irreversible expansion, there are viscous stresses present in the gas that allow the force per unit area at the piston face to drop to the new lower value while requiring that force to match the external force on the outer face.  So the work done by the gas on the piston is equal to the external force per unit area times the change in volume:  $$W = \int{P_{ext}dV}$$  This equation is always satisfied, irrespective of whether the expansion is reversible or irreversible.

Is perfect gas viscous?

no...............

@Anonymous There is no such thing as a perfect gas.  Viscous stresses are a feature of the behavior of all real gases, which the perfect gas concept is incapable of capturing.  And for irreversible expansions, it is viscous stresses that make the difference between the perfect gas equation applying within the gas, and the actual stresses present that enable the compressive stress of the gas on the massless piston to equal the external pressure.  You are aware that, in an irreversible deformation, an ideal gas does not satisfy the ideal gas law, right?  It applies only to equilibrium states.

@ChetMiller I was answering the learner

Consider a thought experiment; imagine playing your favourite sport competitively against someone. Imagine  losing the game as time runs out (bleak scenario sorry). The work is what you actually put in to catch up. If they are good it becomes even harder to catch up. Winning is just how much better you were with your work ethic over theirs. Work is done against the external force. The external force is the resistance to which effort must be put in to overcome it. The external force is your opponents score that you must achieve. Achieving this is work. The resultant force or resultant pressure (internal minus external) merely causes an acceleration and tells you how much time it will take to achieve a particular distance change. If your internal pressure is much larger than your external pressure before equilibrium the faster it will take to reach your desired volume change due to a bigger acceleration. The resultant is just how much better you are over your opponent and this will show by the massive scoreline difference (acceleration) after the game. If the system is quasi-static the internal pressure is only infinitismally larger than the external pressure which means the resultant is infinitismally small which means the acceleration is really tiny, which means you were only slightly better than your opponent because the scoreline was almost tied. Your work, however, is still how much you had to put in to match your opponent therefore the work is based on the external pressure just as work against friction is based on the frictional force and work against an incline is based on the resisting forces of inclination because this is what you need to overcome. In free expansion a gas expands without an opposing external pressure. Even if the original internal gas pressure was 2 or 10 atm, the work is still zero because no opposing force is present. If your opponent doesn't show you cannot call it work. Work is a measure of how much hell someone gives you before you win. So you use their measure not yours. This is why we use external pressure. Resultant force doesn't indicate work, it indicates how easy or hard you overcame that work (or resistance) as measured by an acceleration. In ideal gas behaviour a piston is considered massless therefore resistance by inertia of the piston is Zero. The only resistance is therefore the external pressure and thus this is what you must work against.

The derivation of pressure-volume work is as follows, in reverse:
$W = P\Delta V$
$P = F/A$
$\Delta V = A\Delta d$
Subbing those last two into the first:
$W = (F/A) (A\Delta d)$
Or:
$W = F \Delta d$ which is the definition of work.
So, the work done by the gas in the cylinder is indeed independent of the outside pressure, it is a function of the force exerted by the gas and the displacement of the piston. But, the work done on the piston is different. It is the sum of the work done by the gas inside and that done by the gas outside. The gas outside is exerting a force inward while the distance moved by the piston is outward so that work is negative compared to the positive work done by the gas inside the cylinder. So the gretaer the difference between inside and outside pressure, the greater the net work done on the piston. But still, the work done by only the inside gas on the piston is a function of its pressure and the $\Delta V$.

This is even my confusion that in physics when we write the work done by gas we use its pressure and we don't calculate the work done in irreversible processes because in such processes pressure of the gas at any moment is not defined but in chemistry when there is irreversible process we actually find the the work done by the surrounding on the gas which we take as "-Pext*dV" (-sign is taken to get the right result, you can derive it taking initial volume as the origin and Final volume as end you will get the same result) and reverse its sign to find the work done by gas, as use it in FLOT, for e.g, if there is irreversible compression lets say work done by surrounding on the gas is "+W" i.e W amount of energy is transferred from the surrounding to the system so , it also means that system must also have done -W amount of work on the surrounding to receive W amount of energy  from the surrounding..as there is no other thing from which energy can be exchanged. ...I don't know why we don't do this in physics like we do to find work done in irreversible processes in chemistry, but I think this is not wrong even in irreversible process you cannot calculate work done by gas directly by you can calculate how much you are supplying and it will do the same amount of work on you but of opposite sign to take it from you.


text:
Random order of melting points of group 2 elements

Compare the melting points of $\ce{BeF2, MgF2, CaF2, SrF2, BaF2}$.

Actual order is  $\ce{ BeF2 > CaF2 > SrF2 > BaF2 > MgF2}$.
I have worked out the following things while comparing their melting points.
a) Melting points of ionic compound increases as the ionic interactions between cation and anion in the metal lattice increase. In this case, smaller cations will result in stronger interactions. That gives this order :  $\ce{ BeF2 > MgF2 > CaF2 > SrF2 > BaF2}$.
b) None of these compounds is a covalent solid. If it were, its melting point would have been higher.
c)The order of melting point also depends upon the crystal lattice structure.
(i) Be and Mg crystallise in hcp, Ca and Sr in fcc and Ba in bcc.
(ii) The cations will occupy the lattice points and since fluoride ions are smaller than cations, they will occupy the voids.
(iii) The more is the coordination number, the more will be repulsions between similar ions. In the lattice, when two cations are in contact, there will be repulsions. But, the repulsions will be less when the size of carion will be small since in that case, the electron cloud of one cation will experience nuclear attraction by the other cation.
(iv) Hcp and fcc have the same coordination number.
(v) Based on this logic, the order should be:
         $\ce{ SrF2 > CaF2 > MgF2 > BeF2 }$.
(vi) Although, the coordination number of Ba is very less, it size is
          very big. So, we cannot say with certainty if it would have the
          lowest melting point.
Now, I don't know how to reconcile these two points, the factors are totally opposing. Where to put Barium? This made me think that the order cannot be predicted with certainty just based upon theoretical analysis of their lattice structures.
It is important for me to know whether the points that I have mentioned correct or not. Are there more factors that could influence their melting points? And finally, if the order can be predicted with certainty purely on theoretical basis, how would that be done?

So are you asking about the elements or are you asking about the fluorides?

The fluorides...I understand that the fluorine ions will just occupy the voids made by the metal lattice,..so how does it matter if we analyze the crystal structure of the metal lattice or the metal fluorides. Correct me if I am wrong

related http://chemistry.stackexchange.com/questions/53447/melting-and-boiling-point-trend-in-group-ii

Your assumption that "that the fluorine ions will just occupy the voids made by the metal lattice" seems not to be correct to me.

@aventurin Correct me.


text:
Inability of alkyl fluorides to be reduced to an alkane in the following reaction
Alkyl halides except fluorides on reduction with $\ce{Zn}$ and dil. $\ce{HCl}$ gives alkane. Example:
$$\ce{CH3Cl + H2 -> CH4 + HCl}$$
Why cannot alkyl fluorides be reduced?

Other answers may be more elegant, but it boils down to the carbon-fluorine bond being too strong, in fact stronger than carbon-carbon or carbon-hydrogen bonds.  The required bond breaking simply cannot be done under normal conditions.  Similarly Grignard reagents (formed with a much more powerful reducing agent) are much less common with fluorine than with heavier halogens, and the bad things chlorofluorocarbons do to our ozone layer come from breaking $\ce{C-Cl}$ rather rhan $\ce{C-F}$ bonds.


text:
How to predict which gas shows maximum ideal behavior at STP ? (given its critical temperature)
Suppose we know the Van Der Waals coefficients $a$ and $b$ for a set of gases in the real gas equation, then which is the correct method to predict which gas shows most ideal behavior at STP ?
Our professor said that we must check the critical temperature using the formula $$T_c=\frac{8a}{27Rb}$$ Then he said: "The gas for which $$|T_c-273|$$ is minimum will show the most ideal behaviour".
I sort of don't agree with this logic. Let us have a look at Andrew's curves:

We notice that greater is the temperature compared to critical temperature more is the ideal behaviour. So the correct method should be checking $(273-T_c)$. The gas which has maximum value of $(273-T_c)$ should show maximum ideal behaviour at $273 K$. Is my logic correct or not ? Why?

Its not clear why since a $T_c$ of $173$ and $373$ gives the same result. Presumably he means the one with the lowest $T_c$ in degrees Kelvin.

@porphyrin Yeah that is one problem. But the bigger problem is he said that difference between Tc and 273 should be minimum for ideal behavior. Shouldn't it be maximum for ideal behavior ?

Its just a conversion. Helium seem to have the lowest critical temp, $-268$ C or$ 5$ K, (hydrogen $33$ K, nitrogen $126$ K). Values seem to be mainly listed in centigrade not Kelvin, hence the $273$ is to convert. Can't see what $273$ has to do with it otherwise. Lowest temp (in K) should exhibit most ideal behaviour.

"Lowest temp (in K) should exhibit most ideal behaviour." <--- You mean the gas having lowest critical temperature shows most ideal behaviour at 273 K ? @porphyrin

I think you should read the question. "The most ideal behaviour at STP". If a gas has a high critical temperature, by the time it reaches STP, it would have deviated way too much from ideal behaviour.

If he did not use the absolute value Syntax, he would be correct.


text:
Redox reactions of halides in hexane
When we have two test tubes and we add:

In the first test tube: hexane, aqueous solution of $\ce{NaCl}$ and aqueous solution of $\ce{I2}$
In the second test tube: hexane, aqueous solution of $\ce{NaI}$ and aqueous solution of $\ce{Cl2}$

We are asked to write the reactions that take place in these experiments. Do we write only this:
$$\ce{NaCl + I2 → no reaction}$$
$$\ce{2NaI + Cl2 → 2NaCl + I2 (displacement)}$$
Or should hexane appear in the reactions too?

I'd write the reactions the same way you did. What puzzles me is why they bothered to mention hexane at all.

Hexane is the solvent for the two halogens. It does not react with either so leave it out of your answer.


text:
How does a volcano for a science fair work?
In this video it shows me how to make a Volcano but what is the science behind it? http://www.youtube.com/watch?v=9b_gltKtERY

If somebody wants to answer this, I'll spare you the "trouble" of watching the video - it's a baking soda + vinegar volcano.

This experiment is not to be confused with the ammonium dichromate vulcano experiment.

The food color-dyed water is just there to add volume and color.
The household chemicals used are acetic acid (vinegar, $\ce{CH3COOH}$) and sodium bicarbonate (baking soda, $\ce{NaHCO3}$).
The acetic acid and sodium bicarbonate react to form carbon dioxide gas ($\ce{CO2}$), dissolved sodium acetate ($\ce{CH3COONa}$) and water ($\ce{H2O}$).  It is the liberation of $\ce{CO2}$ gas that causes the foaming and bubbling.
The chemical reaction is written as:
$$\ce{CH3COOH(l) + NaHCO3(s) -> CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)}$$
Where:
l = liquid
s = solid
aq = aqueous (dissolved)
g = gas (vapor)


text:
Possible ammonia anomaly?
In class we have learned that an acid + a base/alkali --> salt + water.
For example:

hydrochloric acid + sodium hydroxide --> sodium chloride + water

However, when I try the same thing with ammonia, I get something like the following:

hydrochloric acid + ammonia--> ammonium chloride

Since there is no oxygen present on the left side, I cannot possibly put water as an extra reagent on the right. However, doesn't this contradict the fact that this type of reaction always forms extra
water? (I do realize there will always be water present, as either the acid and/or the base are in aqueous solution, but I'm talking about the formation of new water) Is this an anomaly of ammonia, or is there something fundamentally wrong about my equations?

Is this an anomaly of ammonia, or is there something fundamentally wrong about my equations?

Fundamentally wrong would be a bit harsh ;)
But notice that the reaction
$$\ce{HCl + NaOH -> NaCl + H2O}$$
is just a special case of a general reaction between an acid and a base. It is special in the sense that the base on the left side of the equation is a hydroxide.
The more general equation for this kind of reaction is
$$\ce{HA + B -> A- + HB+}$$
With other words, the Brønsted-Lowry acid donates a proton to the base. As a result, the acid $\ce{HA}$ is converted to its corresponding base $\ce{A-}$ while the base $\ce{B}$ is converted to its corresponding acid $\ce{HB+}$.
In your first equation, $\ce{OH-}$ is the base, while $\ce{H2O}$ is its corresponding acid.


text:
Classification of Oxides
There is a way to classify oxides according to their behavior where we have the following categories: Acidic, Basic, Amphoteric and Neutral. But is there a way of classifying the oxides according to their chemical bond?

Related: http://chemistry.stackexchange.com/questions/22189/how-to-decide-acidic-or-basic-nature-of-oxides?rq=1

The classification we use to distinguish bonds in other chemical compounds will obviously be valid for the classification of oxides.
They can be classified as covalent or ionic.


text:
How to calculate the final temperature of a gas when it undergoes adiabatic expansion?
The question is as follows:

A sample of $\pu{4.0 moles}$ of a gas ($C_{v,m} = \pu{21 J mol-1 K-1}$) has an initial pressure of $\pu{304.4 kPa}$ and an initial volume of $\pu{20 dm3}$ at $\pu{270 K}$. It undergoes an adiabatic expansion against a constant external pressure of $\pu{79.8 kPa}$, until the volume has increased by a factor of $3$. Calculate the final temperature.

I have calculated:  $Q = 0$  and $w = \pu{-3192 J}$ and change in internal energy $=\Delta U = \pu{-31922 J}$.
The next question is to calculate the final temperature of the gas. However, I am unsure on how to calculate this. Originally I thought as $Q=0$, then the final temperature would be the same as the initial temperature. However, I have realized this is incorrect.

What is the relationship between change in internal energy and change in temperature for an ideal gas?

Rather than answer the question numerically I have outlined the four different cases, reversible / irreversible and isothermal / adiabatic.
In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work done must therefore be at the expense of the internal energy. If the ‘system’ is a gas then  its temperature will not remain constant during any expansion or compression.
In expansion the work done is $dw=-pdV$ and the change in internal energy $ dU=C_vdT$.
The heat change is zero then $dq=0$ which means from the First Law
$$dU = dw$$
and so
$$ C_vdT = -pdV$$
Dividing both sides by T and for one mole of an perfect gas $p=RT/V$ thus
$$C_v\frac{dT}{T}=-R\frac{dV}{V}$$
If the gas starts at $T_1,V_1$ and ends up at $T_2,V_2$ the last equation can be integrated (and rearranged) to give
$$\ln\left (\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
or
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
using the relationship $C_p = C_v+R$
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{(C_p-C_v)/C_v}$$
Using the gas law this van be rewritten in a more useful form as
$$ p_1V_1^{\gamma } = p_2V_2^{\gamma }$$
where $\gamma = C_p/C_v$. This is also written as $pV^\gamma = const$.
The change in internal energy in an adiabatic process is
$$\Delta U=C_v(T_2-T_1)$$
Adiabatic reversible
In a reversible adiabatic change we use the formulas above to work out what happens. If n moles of a gas fill a container of volume $V_1$ at $p_1$ atm. and is expanded reversibly and adiabatically until it is in equilibrium at a final pressure $p_2$ we can calculate the final volume and temperature. The values of $C_p$ and $C_v$ are assumed to be known and are constant with temperature.
The equation to use is
$$ p_1V_1^{\gamma} = p_2V_2^{\gamma}$$
and as only $V_2$ is unknown this can be calculated.  The work done is
$$w=nC_v(T_2-T_1)$$
Adiabatic irreversible
In an irreversible adiabatic change if n moles of an perfect gas expands irreversibly from a pressure of $p_1$ against a constant external pressure $p_2$ the temperature drops from $T_1$ to $T_2$. We can calculate how much work is done and the final volume.
The internal energy change is $\Delta U= nC_v(T_2-T_1)$
and the work done is $w=p(V_2-V_1)$ and as $\Delta U = w$ (as the system is adiabatic) the final volume can be obtained.  Similarly if the volumes are known the final temperature can be obtained.
Isothermal irreversible
In an isothermal irreversible change the work done on suddenly allowing a perfect gas to expanding from $V_1$ to $V_2$ is determined by the final external pressure $p_2$ and is
$$ q=-w =\int_{V_1}^{V_2} pdV = p_2(V_2-V_1)$$
thus expansion into a vacuum does no work.
Isothermal reversible
A reversible isothermal change performs the maximum possible amount of work, and assuming a perfect gas,
$$ q=-w_{rev} =\int_{V_1}^{V_2} pdV=nRT \int_{V_1}^{V_2} \frac{1}{V}dV$$
$$w_{rev} =- nRT\ln\left(\frac{V_2}{V_1}\right)= - nRT\ln\left(\frac{p_1}{p_2}\right)$$

Work done in adiabatic process $W =\frac {nR \Delta T}{1-\gamma}$
$Cv$ (specific heat at constant volume) $= 21 \frac{\mathrm{J}}{\mathrm{mol}*\mathrm{K}}$
$Cv = \frac{fR}{2}  \to f = \frac{21\times2}{8.314}= 5$
So $\gamma= \frac{f+2}{f}$, where $f$ is the degree of freedom.
$\to γ = \frac7 5$
Now work done at constant pressure $= 79.8\times(1000)\times[60(10^{-3}) - 20(10^{13})] = 3192 \,\mathrm{J}$ (positive work done by gas. The internal pressure will simultaneously be equal to the external pressure otherwise first law of thermodynamics won't hold true.)
$W = \frac{nR ΔT}{1-γ}$
$3192 = \frac{4(8.316)(T - 270)}{1-\frac{7}{5}}$
$T = [\frac{3192(-0.4)}{4(8.316)}] + 270$
$\to T = 231.61 K$


text:
Theoritical amount of Sodium Bicarbonate
I have to find the theoretical amount of sodium bicarbonate. The steps I followed in order to produce it are:

15g $\ce{NaCl}$ are added to a beaker. Then we add 87.5 mL of $\ce{NH3}(12M)$
We add 115g of dry ice(whose density is 1.5 g/mL)
Then the solid is taken with filtration and we rinse it first with 9mL of cold water and then with 6mL of acetone.

Then we put the solid in the oven. I know that the product after we take it off the oven will be sodium bicarbonate.
I believe that in order to find the theoretical amount of sodium bicarbonate I have to write down the reactions that are taking place.
$\ce{NaCl + NH_3 + CO_2 + H_2O → NH_4Cl + NaHCO_3}$
But I'm confused as to if there is another reaction where $\ce{NH_4Cl}$ and $\ce{NaHCO_3}$ react with water or acetone. Also, working with the stoichiometry of the reactions, from which reactant is it better to conclude about the moles of NaHCO3?

considering your second question, when looking at the reaction $\ce{2A + B -> AB + A}$, how much AB do you think you get? 1 or 2?

I get 1 AB. But we have to know the moles of A and B in order to see how much reacts from each and how much AB we will finally get.

yeah. So have you calculated how many moles you have of your reactants?

The moles of NaCl are 0.256, the moles of NH3  are 1.05. I'm confused about the moles of CO2(which is produced by the dry ice)

$\ce{CO_2}$ moles are not produced by dry ice. Dry ice is $\ce{CO_2}$.
You need 1 $\ce{NaCl}$ to get 1 $\ce{NaHCO_3}$, right? So if you have 0.256 mol of $\ce{Na}$, how much $\ce{NaHCO_3}$ could you theoretically synthesize (no losses, no by-products)?

Yes I didn't say it correctly, I meant that CO2 is dry ice. Theoretically we could synthesize o.256 mol of NaHCO3. But do we have to take into consideration the moles of NH3? Also, in order to fully conclude that the moles of NaHCO3 are 0.256, we should write the reactions between NH4Cl and NaHCO3 with water and acetone.

For the main reaction you need a 1:1 ratio of $\ce{Na}$ and $\ce{NH_3}$. How would you "take the moles of $\ce{NH_3}$ into consideration"??
Have you checked what reactions between your products and water & acetone can happen? What do you think could happen? Does it even matter? Have you done the reaction? How much bicarbonate have you got?

Since the moles of NH3 are 1.05 and the moles of NaCl are 0.256, it means that NH3 exceeds so my fault. I really don't know if they react with acetone, I just know that we used it in order to dry the solid.


text:
Enthalpy in adiabatic process

A gas adiabatically expanded from $\pu{32atm}$ and $\pu{273K}$ to $\pu{1atm}$ and $\pu{251K}$, Calculate Joule-Thomson coefficient $\mu$ at $\pu{273K}$.

Answer:
$$\mu = {\Delta T \over \Delta p} = {-22 \over -31} = 0.7$$

I think the answer uses the definition of $\mu$ that is $\displaystyle\left({\partial T \over \partial p}\right)_H$, but this definition assumes the process to have constant enthalpy.
But in adiabatic process $dH = dU + Vdp + w_{ad}$, so for $dH = 0$ we need to have $dU + Vdp + w_{ad} = 0$ which is not possible since pressure is clearly changing and so $Vdp$ is not zero.
My questions are

Is $dH = 0$ for a adiabatic and I am missing something ?
If not then how did the author used $\displaystyle\left({\partial T \over \partial p}\right)_H$ to get the answer ?

Do you not understand how adiabatic flow through a porous plug or valve, from high pressure upstream to low pressure downstream, gives no change for the enthalpy change per mole?  Are you aware of the equation:  $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

No I only know $\displaystyle dH = -\mu C_pdp + CpdT$.

If you combine these two equations, you get a relationship for calculating $\mu$ from knowledge of the heat capacity and the equation of state.

Ok I subtract my equation from our equation,  $$0 = \left[V-T\left(\frac{\partial V}{\partial T}\right)_P +\mu C_p\right]dP$$
How can I get rid of $C_p$ ?

You can't.  But, in any event, you did the algebra incorrectly.  What happened to the dT.

$$dH - dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP -  ( -\mu C_pdp + C_pdT) =C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP +\mu C_pdP - C_pdT  = \left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP +\mu C_pdP = 0$$
can't I just cancel $dT$ ?

You want to be solving for dT/dP at constant H.

Since $H$ is constant I get $\displaystyle 0 = -\mu C_pdP + C_pdT$ or  $\displaystyle {dT\over dP } = {C_p\mu\over C_p  } = \mu$. But I did not use your equation ?

$$\mu=-\frac{\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]}{C_p}$$

@ChesterMiller But how is that useful ? since I don't know $V$ and $C_p$.

If you know the equation of state of the gas (P,V,T) and its heat capacity at constant pressure (which is also often known), you can predict in advance the JT coefficient.  But, that wasn't your question, and I was trying to address your question.  Regarding how he got his answer:  He just measured $\Delta T$ and $\Delta P$, and thus determined $\mu$ experimentally.

In this problem, calling $\Delta T/\Delta P$ the Joule-Thompson coefficient really isn't proper for such a large change in P .  The Joule-Thompson coefficient $\mu$ is, strictly speaking, defined only in terms of an infinitesimal difference in pressure between the upstream and downstream sides of the porous plug (and a corresponding infinitesimal change in the real gas temperature).  A change from 32 atm to 1 atm does not really qualify.  At best, the data provided would give a very crude approximation to $\mu$.  And the equations that we have given in terms of Cp and the EOS (continued)

(continued) would not give an accurate prediction of $\Delta T/\Delta P$ for a typical real gas.

The Joule-Thomson experiments occurs with no change in enthalpy.
Suppose that at the left of a porous plug there is a pressure $p_1$ and temperature $T_1$ and $p_2,T_2$ to the right of the plug, as $p_1>p_2$ the gas moves left to right. The experimental configuration must ensure that pressures remain constant and that the experiment is performed under adiabatic conditions when $q=0$.
If a volume $V_1$ of gas moves from the left to right the work done/mole is $W=p_1V_1-p_2V_2$. This is the difference between the work of compression on the left of the plug and work recovered on expansion on the right. If the gas were ideal then $w=0$, but real gases are not. The gas expansion is also adiabatic so that no heat leaves or enters then $q=0$ and the change in internal energy $\Delta U$  is equal to the net work $$\Delta U =U_2-U_1 = p_1V_1-p_2V_2$$ therefore $$U_2+p_2V_2=U_1 + p_1V_1$$
As $H=U+pV$, then  $$\Delta H = H_2-H_1=U_2 +p_2V_2-U_1 -p_1V_1 =0$$
The Joule-Thompson  coefficient $\mu$ is defined, as you write, $\left ( \partial T/\partial P \right)_H$ and this measures how much the intermolecular interactions make the gas differ from an perfect gas. Most gases cool when passing from high to low pressures at room temperature.
Notes:
The coefficient can be rewritten in other forms using
$$ \left ( \frac{\partial T}{\partial p} \right)_H \left ( \frac{\partial H}{\partial T} \right)_p \left ( \frac{\partial p}{\partial H} \right)_T =-1$$
then
$$ \mu C_p= -\left (\frac{\partial H}{\partial p}  \right)_T $$
as
$$ \frac{dH}{dP} = C_p\frac{dT}{dp} +V-T\left (\frac{\partial V}{\partial T}  \right)_p$$
then if $\alpha=(1/V)(\partial V/\partial T)_p$ is the coefficient of expansion then at constant T, $(\partial H/\partial p)_T= V(1-\alpha T)$ which means that the Joule-Thompson coefficient can be written as $\mu=(V/C_p)(\alpha T-1)$ (This last equation has been used as a way of measuring absolute temperature because $V,\mu$ and $ \alpha $ are all measurable quantities.

Most gases cool when passing from high to low pressures at room temperature?  Not according to Wikipedia:  https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect

perhaps 'most' was loose, 'many' would be better, or best only  those with positive coefficients at the prevailing temperature will cool, i.e. $\alpha T>1$

@porphyrin What is the $p$ in the partial derivative? $p_1$ or $p_2$? Why for ideal gas $w=0$? Thanks in advance.

@adosar corrected typo in $\Delta H$  eqn. The last equations are general for any pressure.

@porphyrin So we can measure either the change of $T_1$ with respect to $p_1$ or $T_2$ with respect to $p_2$?

@ChetMiller  I believe that Poutin is correct for most real-world applications, i.e., those in which p_final ~ 1 atm, and p_initial is not more than ~ a few hundred atm.  In that regime, the only gases I know of which heat during a JT expansion at room temp. are hydrogen, helium, and neon. Do you know of any others? The interesting question would be what happens when the initital pressure is so high that it puts, say, N2 into a heating regime at room T -- then as its p decreased it would first heat and then cool; you could test or calc. how high p_initial would need to be for heating to win.

@theorist Why don’t you start a new thread so that we can analyze this in greater detail?

@ChetMiller That's a nice idea, but in posing the question I'd first have to see if I could derive an expression for the pressure-depence of the JT coefficient at v. high p for a specific gas. [Once I had that, I could integrate to get $\Delta T$.] I'd also need to account for a possible transition from gas to liquid or SC fluid at v. high p, which would in turn require a different expression be obtained. [Plus I'd need to think about the effect of a phase transition within a JT expansion upon $\Delta T$!] All of this requires time, so it will need to wait until I have a chance...

@theorist I think all this is doable, depending on the kind of data one has available, including large pressure drops (not just differential pressure drops).


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

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

Going to vary depending on which oxidant you choose

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

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

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

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

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


text:
Final temperature of a adiabatic process

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

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

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

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

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

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

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

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

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

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

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

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

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

288 is correct.

@ChesterMiller Yes thanks.  Can you write an answer ?

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

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

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


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

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

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

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

@Tyberius Second answer solved my problem. Thank you.


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

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

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

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

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

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

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

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

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

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

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

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

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