text:
What exactly is the Horwitz equation?
During a laboratory exercise in a course on analytical chemistry, we were asked to calculate the $z$-score.

$$z=\frac{X_{\text{lab}}-X_{\text{cons}}}{S_{\text{target}}}$$
  where

$X_{\text{lab}}$ is some measured quantity (e.g., concentration),
$X_{\text{cons}}$ is the consensus value for that $X_{\text{lab}}$,
$S_{\text{target}}$ is known as the benchmark standard deviation

One way of optaining the $S_{\text{target}}$ was to use the so-called Horwitz equation:

$$S_{\text{target}}=2^{(1-0.5\log C)}.$$
  $C$ denotes the concentration as a dimensionless mass fraction.$^{[1]}$

On the mathematical relationship,  M. Thompson comments$^{[2]}$

As well as being useful, the Horwitz Trumpet is a feature of considerable theoretical interest. It is hard to avoid the assumption that a simple mathematical law that describes the behaviour of large numbers of methods over at least six orders of magnitude of analyte concentration must have some inherent meaning and deserves serious consideration. So far,
  though, nobody has managed to explain the strange empirical exponent from basic principles, although several people have made conjectures. Are we seeing the manifestation of a physical law here, or is there a psychological basis, perhaps to do with our perception of fitness for purpose?
There is a sure-fire paper in Nature waiting for somebody!

In statistical terms, why have we chosen such a form for $z$?

This might include an explanation of the relationship with the Horwitz equation.

What are the several conjectures that have been made?

[1] Carlos Rivera, Rosario Rodríguez. Horwitz Equation as Quality Benchmark in ISO/LEC 17025 Testing Laboratory
[2] M. Thompson. The amazing Horwitz function. Royal Society of Chemistry. (2004)

Look i am from mexico, so my English is not perfect but horwitz was the editor of the AOAC that is a book that contains test methods for various matrixes, during this time Horwitz recouncade data of validations of these methods and obtained an estimate of the behavior of the variability of the results and described it in the AOAC with the equation you have, so that the result of that equation says the estimated dispersion of the values with respect to your concentration of the analyte, for more information checks the appendix of validation of the AOAC. I will review the site conditions and if

I think I agree with M. Thompson that the Horwitz equation is useful but disagree rather intensely with the idea that is is "of considerable theoretical interest."

The Horwitz function 1, also known as the Horwitz “Trumpet”, is an empirical relationship that has, thus far, not been derivable from fundamental principles. As noted by Thompson 2, and as shown below in Fig. 1 (from 2), the relative standard deviation of reproducibility (RSDR) increases as analyte concentration decreases:

Horwitz formulated his expression after noticing a pattern in the relative standard deviations reported in many (thousands at this point) collaborative trial results. The following figure (a screenshot from Hanley [3, p. 12037]) provides the essential equations and background information:

As Hanley notes, the Horwitz function appears to be in the Tweedie family of exponential dispersion fluctuation scaling relationships (https://en.wikipedia.org/wiki/Tweedie_distribution ). Specifically, the Horwitz function may be a compound Poisson-Gamma process, since the exponent, $\alpha $ in equation 2, is approximately 0.85. (More accurately, Thompson 1 notes that it is $1 - (log2)/2$).
Answer to OP’s first question
Assume measurements are Gaussian distributed, with $\mu $ as the population mean and $\sigma $ as the population standard deviation, and assume both parameters are unknown. Let $X_{lab} $ be a single measurement, i.e., a random sample from the Gaussian distribution. Then $z = (X_{lab} - \mu )/\sigma $. So z, which is the so-called ‘z score’, is simply the signed difference parsed in units of $\sigma $. A small z score means the measured result is close to the true value, i.e., $\mu $, assuming no systematic error. A large z score means the measurement has relatively poor accuracy.
But both $\mu $ and $\sigma $ are unknown, so $X_{cons}$ is used as an estimate of $\mu $ and $S_{target} $ is used as an estimate of $\sigma $. The result is a rough, and possibly biased, estimate of z. Frankly, this is too crude for me.
Answer to OP’s second question
Also as noted by Hanley, attempts have been made to derive the Horwitz function from “the binomial distribution, Zipf’s law, and a heuristic approach”: see 4 and 5 below. However, these have not been generally been accepted and it is fair to say that the final word has not been written on the origin of the Horwitz function. As well, the Horwitz function potentially has a problem at zero analyte concentration, as noted by Holland et al. [6]. This is problematic because analytical blanks do not have infinite standard deviations.
References:

W. Horwitz, “Evaluation of analytical methods used for regulation of foods and drugs”, Anal. Chem. 54 (1982) 67-76.
M. Thompson, “The amazing Horwitz function”, AMC Technical Brief No. 17, July, 2004 Royal Society of Chemistry. (2 pages).
Q.S. Hanley, “Chemical Measurement and Fluctuation Scaling”, Anal. Chem., 88 (2016) 12036-12042 (Open Access).
R. Albert, W. Horwitz, “A Heuristic Derivation of the Horwitz Curve”, Anal. Chem., 69 (1997) 789−790.
P. Hall, B. Selinger, “A statistical justification to relating interlaboratory coefficients of variation with concentration levels”, Anal. Chem., 61 (1989) 1465−1466.
R. Holland, R. Rebmann, C.D. Williams, Q.S. Hanley, “Fluctuation Scaling, the Calibration of Dispersion, and the Detection of Differences”, Anal. Chem., 89 (2017) 11568-11575.


text:
How does osmosis work?
I have (very) basic knowledge in chemistry.
I am curious as to why osmosis works. I read a couple of introductory chemistry books and also searched online but most explanations I found only explained the phenomena (i.e. solvent molecules move through semipermeable membrane from low solute concentration area to high concentration area), but not the reason behind it at the molecular level (i.e. what makes more solvent molecules move towards the high solute concentration area than in the reverse direction).
Wikipedia seems to go a step further:

... there is an interaction between the solute and water that counteracts the pressure that otherwise free solute molecules would exert.

And a bit later:

The virial theorem demonstrates that attraction between the molecules (water and solute) reduces the pressure, and thus the pressure exerted by water molecules on each other in solution is less than in pure water, allowing pure water to "force" the solution until the pressure reaches equilibrium.

I think I understand what it means, but what I don't understand is this: osmotic pressure is said to be a colligative property, i.e. only affected by the concentration of the solute and not by its type. If the reason for osmotic pressure is the attraction between water and solute molecules - wouldn't we expect different osmotic pressures for different kinds of solutes?

@obe try to think about entropy changes, third law of thermodynamics and remember the fact that in osmosis solutes do not cross membrane.

@JM97 honestly I don't know much about thermodynamics. I did read about it a little. I read that a system always moves towards a higher entropy state. The explanation didn't go into the low-level mechanics but the way I imagined it was that since free particles move randomly within a system, they are bound to spread out evenly (on average), thus increase the system's entropy. If this physical description is correct (is it..?) - then it makes sense to me that solvent molecules should still go through the permeable membrane at the same rate in both directions... What am I missing here?

@JM97 yes, but how come the interaction doesn't depend on the type of the solute? Wouldn't the interaction be different depending on the polarity / size / geometry / etc... of the solute?

@JM97 but "why"? why is there a net flow from pure water into the solution?

Let us continue this discussion in chat.

It does depend on the type of the solute, but it doesn't matter much in the big picture. The main thing that matters is that you have a polar solvent, and you dissolve a polar something in it - the change in the solvent means that there's a net "lack of solvent" on the side with the solution, which manifests as a (thermodynamic) pressure. I think you're missing the point of your quote - the solution is not just a mix of a solvent and something. Why doesn't it matter what the solute is? Because solvent is the important bit - the solute "steals" "water molecules"; only concentration matters.

there is a answer here https://chemistry.stackexchange.com/questions/167189/osmotic-pressure-definition-clarification/167286#167286 describing how Osmosis works as a molecular level

According to this document (https://arxiv.org/pdf/physics/0305011v1.pdf)

"Water molecules can pass through the membranes in either direction, and they do. But because the concentration of water molecules is greater in the pure water than in the solution, there is a net flow from the pure water into the solution."
  Hydraulic equilibrium is achieved when the lower concentration of water in the high pressure side is balanced by the higher pressure and therefore higher energy pV on that side

This is the diffusion model of osmosis which is easy to understand at your level.

Thanks! This paper looks very relevant. I'll try to read it and come back...

I don't think that your answer answers my question - it doesn't describe the process at the particle level but rather uses derived concepts, however the article you linked to is the most relevant response I received for this question. However, while the article acknowledges that osmosis is not properly explained in a qualitative manner in chemistry texts, I am not sure it itself does it properly. It seems to go a bit further in the qualitative description but then jumps to equations instead of following through. Or maybe I just lack sufficient background to thoroughly understand...

To first order the solute doesn't matter, though it probably has a small influence. The driving force for osmosis is entropic. The osmotic pressure then is not due to the forces between the solute and solvent, but mostly from the fact that you can mix the two together.
Indeed, osmosis works even when there's less interaction between the solvent and solute, but at some point, the solute is no longer soluble in solvent, so there is no mixing.

Can you explain the forces that are exerted on solvent molecules that cause more of them to move or stay to or at one side of the membrane?

I think you're looking at this wrong. Imagine a container that only had gas molecules in one half it. What force causes the gas to diffuse into the other half? There is no force, but the molecules are moving, and there is a greater likelihood that they will occupy the whole box than just half of it. Well, osmosis is a kind of diffusion driven by the entropy of mixing. We can describe the fact that the process is favorable by defining a chemical potential, but there is no force per se that is operating.

I actually mentioned my understanding of the phenomena of gas filling up the whole box in a comment above. It is my understanding that free particles are in constant movement in random directions, and as such it makes sense to me that they will fill out a container evenly. I admit I have some open questions about the constant movement of particles as well but let's take it as an axiom for now. I still don't see how it explains osmosis. With a membrane that is not a barrier for water, according to the analogy, and ignoring interactions with the solute, I'd expect water to spread out evenly

See, that's the issue. It does spread out evenly, but you're using the wrong definition of evenly. A simplistic way to think about it is that when you add the solute, "evenly" requires that you need more water to make it look like the other side that doesn't have solute.

@Zhe is correct in writing that it is the increase in entropy that drives the process. Entropy here means the increase in probability (number of possible arrangements) which arises from distributing particles throughout a bigger rather than smaller volume.  As the solute becomes diluted its entropy increases.

Two factors are involved. First: the Gibbs free energy or chemical potential of the molecules. This is a function of the temperature which results in their random velocities and the entropy of the molecules, their concentrations.  With no physical constraints and ignoring solvation effects and chemical reactions diffusion occurs until the solution is homogeneous. Entropy is maximized at constant energy; the free energy is minimized.
Second: the semipermeable membrane. These can be natural or synthetic and work on two general principles: The membrane either has holes small enough to sort molecules or a chemical composition that is selectively soluble to specialized chemicals, with possibly a combination of both. An example is one's skin. Skin is hydrophobic with sweat glands to pass water and can be very permeable to small organic solvents.
Osmosis involves two solutions with different chemical potentials and a membrane selective to one or more components of the solutions and impermeable to others. [The selectivity can be very high but is not absolute there is an efficiency.] The solutions are separated by the membrane. The permeable molecules diffuse thru the membrane at rates determined by their chemical potentials until the chemical potentials are equal and the free energy change is zero. The equality can be equal concentrations or equal collision frequencies and can be influenced or hastened by gravity or a centrifuge or by temperature or electrical forces.
If the solute reacts minimally with the solvent, ie. a reasonably ideal solution, osmosis is dependent on the chemical potential, activity, of the permeable substance and reasonably independent of the nonpermeable substances.

I think you can find what you are looking for by googling "entropy"
http://www.mpcfaculty.net/mark_bishop/solubility_entropy.htm
To summarize the article : "In general, particles of matter tend to become more dispersed (spread out)."
My guess is that it has something to do with the leonard jones force. Here is a video I made where the "text" is like a positive force and all the particle are a negative force. So the particles are slightly attracted to the text but when they get too close to the text, they get launched away (because I made the repulsive force to be huge).
-
But in reality every single particle would have a positive and a negative so they all are attracted to each other but push away from each other once they get close enough. https://vimeo.com/182775330

I read the article. Interesting, thanks :) I agree that entropy seems to be key here. However, I still have difficulty with this. I think my main issue is natural tendency toward greater dispersal. I understand why it's the case with gas, at the atomic level, and why it's the case when mixing water and ethanol. I imagine the particles and their random movements and I can understand why they disperse. However applying this to osmosis is a leap I don't understand because when I imagine solvent molecules moving freely, they end up with the same amount on both sides of the membrane...

On one side, the atoms are repelling each other more because they are concentrated. But after they have the same amount on both sides, the pressure is equal....I imagine.

@obe It's in the wikipedia quote you have in your question - when you dissolve something in a solvent, it changes both the something and the solvent. The water no longer behaves entirely like water, which in this scenario means that the side with solvent "feels" like it has less water than it really does. Read up on how solutions (or e.g. acids) work - it's not just a mechanical mix of two chemicals. Think about a salt solution - pure water is an isolant, pure salt is an isolant, but dissolve some salt in pure water, and you get a great conductor - both are "torn apart" in a solution.

@Luaan I realize that but I'm having trouble with the "feel" concept... if I believed water could "feel" things, I'd post on homeopathy.stackexchange.com :) jokes aside, having learnt a little chemistry, I now understand the molecular forces that underlie various macro phenomena. For example, I can explain why water is liquid in room temperature while oxygen is gas, using the intrinsic properties of the nucleus and its sub-particles. I'd like to be able to do the same with the osmosis phenomena...

@obe You know how when you dissolve salt in water, at some point, the salt stops dissolving? At that point, the water "feels" like there's no water - the solution is saturated. I hope you have no problem with this concept? Why is that so? The molecules of water that are already engaged with a solute cannot bind with another molecule of a solute, so there is less water available for solving the solute. This manifests as thermodynamic pressure across the membrane. And you're quite confused - only electrons matter for your example. Nuclei only matter in so far as they pair with those electrons.

@Luaan as far as I know the explanation for saturated solution is that the reaction works in both ways (i.e. solute dissolved and precipitates at the same time). The solution is said to be saturated when the rate of the reaction in both directions is the same. The exact point when it happens depends on several factors such as polarity, geometry, temperature, etc... It can be explained in these terms without using words like "feels"... can you explain how this manifests as thermodynamic pressure across the membrane?

@obe You've explained it yourself. How can you not see it? I give up :)

As far I as I understood from my professor
Osmosis is the flow of water from higher solute concertation to lower through a semipermeable membrane
That solely because of increased collision between water and solute in the high solute concentration area
Now the question arises why water and not the solute movement. Solute being less in size than water technically should move. But here comes thdy(thermodynamics) in play
The let's say the solute here is salt. That when added in water dissociates into Na+ and Cl- now in order to nullify the charge and satisfy the entropy loss the water arranges it self around Na+ in hydration shells this hydration shell helps in dispersal of the +ive charge carried by the Na+ and similar hydration shells are made with Cl- to it's only the orientation of H and OH differs
Na+ makes hydration shell with OH oriented outwards while Cl- with H of water oriented outward.
So the hydrated Na+ and hydrated Cl- are to large to pass through the semi permeable membrane and that's is why it's the water..
Hope it helps. Feel free the correct if sth is wrong

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text:
How to determine the valence electrons of D-block elements in their mono atomic form?
For s and p block we can determine the valence electrons easily,but what about d-block elements. How to determine their valence electrons?

Based on the Aufbau principle, they would all be filling the (x-1)d sublevel, so they would ALL have 2 valence electons in the xs sublevel.
Since the Aufbau priniciple fails to account for the fact that some elements will be more stable with incomplete valence xs sublevels and additional electrons in the (x-1)d sublevel, you just need to memorize the transition metals that are more stable with 1 valence s electron:
Chromium, Possibly Nickel, Copper, Niobium, Molybdenum, Ruthenium, Rhodium, Silver, Platinum and Gold,
And the two extra specials:
Palladium which prefers to fill the x-1d sublevel and not put any in the xs sublevel.  It has 18 in its highest energy level, level 4, but only the 10 4d electrons are considered to be valance electons.
And Lawrencium which has 7s2 (5f14) xp1, or 3 valence electrons, putting 1 electron in the xp sublevel rather than in the (x-1)d sublevel.  It turns out to be a d-block element that is not filling the d-block.
Notes
This will give you the number of electrons in the highest energy level for transition metals.  Transition metals in the highest d sublevel (but not in the highest primary energy level) can provide valence.  Many sources will refer to all of the electrons above the configuration of the prior noble gas to be valence electrons.
On one hand, most of the oxidation states for transition metals have been observed, up to the point where they would have lost all of those d electrons.
On the other hand, the d electrons in transition metals are not de-localized in metallic bonding of transition metals (to the best of my knowledge).

You are right on the partial occupation of orbitals. However, the definition of valence electrons for transition metals is not quite the same as for main group elements. The energies of the d and f orbitals are sometimes close to the highest $n$ orbitals. As a result, they can be considered as "valence electrons". So for d and f block elements, you also have to consider the energy of the orbitals in addition to the electron configuration.
A rule of thumb that I found in wikipedia: the valence electrons in d-block elements are all electrons that are outside a noble gas $e^-$ configuration.

Agree, but I would say that the correct answer will depend on your definition of valence electrons.  The same source will even define valence electrons differently in different situations.  http://environmentalchemistry.com/yogi/periodic/Cu.html

For example, they ^ refer to all of the electrons above the noble gas configuration as being valence electrons, but later state that Cu I has 1 valence electron and Cu II has two "Cu I (cuprous) has one valence electron and Cu II (cupric) has two valence electrons" so you could say that it has as many valence electrons as are providing valence in a given compound.  For example, one could argue that the d electrons become valence electrons after the s electrons are removed.

By the way, newer electron configuration notation puts the xs higher than the (x-1) d for transition metals because IN transition metals they are higher energy and they ionize first.

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text:
Ideal Gas Law: What is P?
I know P is the pressure, but is it the outside pressure or inside pressure? I know that in equilibrium, it doesn't matter, but then in this problem I got confused:
193 mL of O2 was collected over water on a day when the atmospheric pressure was 762 mmHg. The temperature of the water was 23.0 o C. How many grams of oxygen were collected?
How would you do that?

It is the pressure inside the volume of trapped gas, and also the pressure of the surrounding atmosphere (neglecting the tiny hydrostatic pressure difference between the trapped gas and the atmosphere).


text:
Does the salinity in seawater affect the process of alkaline hydrolysis?
I am an architecture student doing a project on crematorium spaces. I am looking at resomation or alkaline hydrolysis as an alternative method of disposition of a body. From what I gather from research, the funeral parlours offering the resomation service uses 80 gallons/ 300 L of water for each body. I am just wondering if in theory such an enormous amount of water could be replaced with seawater? Or does the salinity or pH of seawater inhibit the process?

Interesting! This is not an answer, but there's a master thesis on the Environmental impact of funerals that you might find useful.

What concentration of hydroxide does the job?

The average pH of sea water is approximately 8.  Thus, it is alkaline to begin with, so I don't see how it would be detrimental to replace typical water with sea water in the process described. Further, the salinity should not adversely affect the interaction of lye with tissue, as far as I can determine.  Thus, in theory, one should be able to replace water with sea water in this context.

I believe that desalination would be necessary as the salinity of the seawater would interfere with the earth metal salts already utilized to perform the process of alkaline hydrolysis.


text:
Oxidation number vs charge?
From what I understand:
*oxidation numbers: are the "act" of electrons moving
while
*charge: is what the atom "wants" to do with its electrons.
-
Can anyone please clear up the distinction between the two and when its appropriate to use one, the other, or both at the same time?
-
Is it that charge is only used for ions?

If an atom loses electrons, it develops a positive charge and becomes a cation. Likewise, if an atoms gains electrons, it develops a negative charge and becomes an anion.
"Charges only EXIST on IONS."
Covalent bonding rose the oxidation number.
We are aware that in a Covalent Bond sharing of electron takes place.
The more electronegative atom will take more bonding electrons and it will get a negative oxidation number (It depicts Anion).
Similarly less electronegative atom mimics a Cation.
In other words you can say Oxidation Numbers are fictional ionic charges

Im guessing that means I need to use one or the other, never both at the same time?

Yes, electrovalent bonds carry Charge while covalent bonds carry Oxidation Numbers.


text:
Are d electrons ever delocalized in metallic bonding
It has been my understanding that delocalized electrons that participate in metallic bonds come from the highest primary energy level, specifically from the s and p sublevels and that electrons in d orbitals (from one level lower) do not delocalize.
Palladium however, has its $5d^{10}$ electrons IN its highest energy level because it has no 6s electrons, and yet Palladium clearly has electron delocalization.  So do d sublevel electrons delocalize in general?  In any metal elements? (such as in Palladium).

d-block elements bond with s and d electrons not p.

I didn't say d-block.  I included p because Lawrencium has a p valence electron.  Sources I have found say that d do not dissociate in metallic bonding but they obviously do in Pd.


text:
Why do all inter-halogen compounds contain an even number of atoms?
Why do all inter-halogen compounds contain an even number of atoms?
All halogens have one electron in their valence shell. One covalent bond is formed by 2 halogen atoms (sharing one electron each). If in the compound an odd number of atoms were involved, then an odd electron compound (unstable) would be formed. IF4 would contain 1 lone pair and 1 odd electron, hence unfavorable.
Is my answer correct? If you have any better explanation, please share it.

One unpaired electron.

Yes neutral $\ce{IF4}$ would be unstable because it would have one lone pair and one unpaired electron.  The extra lone pair is not a problem because Iodine is large enough to exceed an octet, but the unpaired electron makes it a free radical.


text:
What is meant by the numbers on reagents above and below arrows?
I was taught that molecules on top of arrows are reactants and below arrow are solvent.
But I've come across a series of molecules written above and below arrows with numbers and commas. Do these numbers mean that all these reagents above and below arrows are in separate containers and are added one by one?

When you have numbers it means that these are sequentially added. So, in this case, you'd add Hg(OAc)2 and HNuc to your starting material, let it react for a certain period of time, then after that add NaBH4.

Is that for all cases where u see the numbers? My TA was saying sometime, it could also mean that all these molecules above and below arrows are dumped together in one container and let to react on their own? I don't know if he was right in saying that?

If there are numbers present, I would say they are always sequentially added. If there are no numbers present, then it's usually added together.

The numbers indicate steps. The only problem is that the definition of a "step" can be slightly different. Sometimes, separate steps mean that you do something, wait some time, and then do the next thing. Other times, you need to work-up or destroy the excess reagents that are left and isolate the desired product before you can proceed with the next step.
In your example, you need to keep the mercury and the borohydride separate. If you used any excess in the first step, you might need to destroy it before you add the borohydride.


text:
Is there a coating substance that is very visible through a lens?
I'm trying to find a way to easily spot golf balls, and I was looking for some chemistry help. I was wondering if there is a substance that I could coat a golf ball in, that would be very visible through a specific type of lens?

A fluorescent paint, with perhaps additional UV illumination. If in sunlight, a few flashes might be needed, with UV-transmission filters on them.


text:
What is meant by acid catalyzed hydration of alkene?
My teacher told me that acid catalyzed hydration needs acid because h20 is a weak acid and conjugate base of OH- is unstable.Can anyone tell me exactly how this is sure in terms of equilibrium?I don’t know the pka of carbocation formed to compare with acidity of H20 and see where equilibrium lies.

This is a bit tough to answer because your question indicates confusion with multiple concepts.
First, you don't measure the $\mathrm{pK}_{a}$ of a carbocation. It is a Lewis acid, but that's not the correct context for this problem. It is an electrophile. It will accept a nucleophile, which in this problem is water (the lone pairs on the oxygen make it an OK nucleophile). The resultant product, an alcohol, has the appropriate functional group (OH) because it comes from the water molecule that is added.
Second, the slow part of this reaction is generating the carbocation. In order to do so, you need to add a $\ce{H+}$ to the alkene. Unfortunately, water is too weak of an acid to do this efficiently. Instead, we use a strong acid, which in water, forms hydronium $\ce{H3O+}$. This is a strong acid and able to protonate the alkene to create the carbocation. As discussed above, the cation goes on to react water to generate the alcohol.
One additional point is that after adding water to the cation, we actually don't have the alcohol directly; we have the oxonium ion: $\ce{ROH2+}$. In order to generate the alcohol, we need to remove $\ce{H+}$. The hydrogen ion goes back onto a water molecule to reform the hydronium ion that we used to start the reaction. By regenerating the hydronium, the reaction is catalytic in hydronium, i.e., it's not used up in the reaction (and technically, it makes the reaction go faster). That's why this reaction is acid-catalyzed.
I strongly recommend going back to review the mechanism and trying carefully to identify which role each of the components plays in this reaction.


text:
What is the problem with my desiccant experiment?
Ok, so here's some context. For my experiment, I wanted to see which desiccant would absorb the most water in a low humidity environment (30% RH). To conduct this experiment, I placed the desiccant into a box with a saturated salt solution (Magnesium chloride) and then placed the box into an incubator to get a constant temperature. Due to practical restraints (and poor planning), I was only able to fix the mass of desiccant to 10g and not the moles of desiccant. I also left the experiment for over 12 hours.
I expected by results to be similar to results I found on the internet but it was nowhere close. My results were:

Silica Gel: 5.84% of its mass absorbed
Calcium Chloride: 7.77% of its mass absorbed
Calcium Sulfate: 6.87% of its mass absorbed
Lithium Chloride (aqueous): 3.02% of its mass absorbed

I'm not sure why my results are so inaccurate I guess. Is it because I didn't use enough desiccants, or is it because I didn't have a direct air flow? I really can't figure out why my experiment is inaccurate....


text:
If a molecule has both polar and ionic bonds in it, is it polar or nonpolar overall?
To be more specific, lets take a look at the Lewis Structure of two such examples:
The first one is SiCl${_3}$F. The red arrow represents an ionic bond while the blue is a polar covalent bond.
Now, the polar bonds on the Cl atoms that are separated horizontally cross out. Looking vertically, however, the ionic bond is stronger than the polar bond, so does that mean the molecule over-all is a polar molecule? I say this because my textbook says it is a tetrahedral polar.

The next one is SiF${_2}$Cl${_2}$. Same here; the ionic bonds and polar bonds should cancel out and the molecule would be a non-polar molecule overall, but my textbook says it is a polar molecule for some reason.

There's no ionic bond in here. Difference in standard electronegativities is not enough to tell, even misleading here.

Beyond what @Mithoron said, remember that these molecules won't be planar -- they're tetrahedral. So, e.g., the 'horizontal' $\ce{Si!-!Cl}$ bonds don't cancel each other out -- they'll still have a residual contribution to the overall polarity of the molecule.

1) The silicon to fluoride bond is polar covalent. Even though an electronegativity difference of more than 2.0 generally indicates an ionic bond, the cutoff rises as the average electronegativity of the two combining species rises.
For example, in sodium bromide, the electronegativity difference is only 1.9, but the average electronegativity between sodium (0.9) and bromine (2.5) is 1.7 which is on the low side meaning that the cutoff will can be lower than 1.9.
In the silicon-fluorine bond, the electronegativity difference is 2.2 which would indicate an ionic bond, but since the average electronegativity between silicon and fluorine is (1.8+4.0)/2 or 2.9, which is on the high side, the threshold for electronegativity difference to form an ionic bond is higher.
For a molecule to be polar, it has to have polar covalent bonds, but not every molecule with polar covalent bonds is polar.  It won't be polar if the molecule is symmetrical across the x, y and z axis cutting through the central atom.  So the first example is polar because it is not symmetrical.
The second molecule is polar because it is also not symmetrical, since it it not square-planar, but rather tetrahedral.

(image source)


text:
When volume decreases and the product has less moles than reactants, why is there no increase in the reaction quotient?
Assume a reaction:
$$\ce{CO + Cl2 <=> COCl2}$$
My textbook says that a decrease in volume will result in the reaction quotient being less than the equilibrium constant. Why is that?
With a decrease in the volume, a forward reaction proceeds. Thus, more product and less reactant means there should be a higher reaction quotient.

Keep in mind that for reaction
$$\ce{r_{1}R_{1} + r_{2}R_{2} + \ldots + r_{n}R_{n} <=> p_{1}P_{1} +\ldots + p_{m}P_{m}}$$
the reaction quotient is:
$$Q = \frac{\prod_{i = 1}^{m} [P_{i}]^{p_{i}}}{\prod_{j = 1}^{n} [R_{j}]^{r_{j}}}$$
Specifically, here:
$$Q = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$
For your specific reaction, the reaction quotient is getting smaller because the denominator has a quadratic dependence on the pressure where as the numerator only has a linear dependence on pressure.
The goal as usual is to make $Q = K_{\mathrm{eq}}$, which is equilibrium.
Since now, $Q < K_{\mathrm{eq}}$, you want to push the reaction forwards. That will increase the concentrations of product and decrease the concentrations of reactant, making the reaction quotient bigger.
Intuitively, that makes sense in your example. There are two molecules combining to form one molecule. By decreasing the volume and increasing the pressure, you're squeezing them together into the product.


text:
Is there a treatment for distilled water to become drinkable?
It is my understanding that, drinking distilled water may not be recommended. Its not recommended indeed. Or maybe its not not recommended... Maybe not not indeed........ Anyway....
Well, apparently, in some places, they actually acquire distilled water from various sources (desalinization, atmospheric moisture, etc), and sell for people to drink.
My question is: What are the methods used to treat distilled water, such that it becomes more like the "tap water" we drink? (or, perhaps, drinkable/safer in the sense of the not-recommended-links (that is, assuming such links are true ofc)).

Drinking distilled water in the course of consuming a normal diet is harmless. Your "not not" links are correct. That said, I'm voting to close b/c personal health questions are off-topic for the site.

@hBy2Py I don't think this is personal health. I was just elaborating in the motivation of the question. I was very clear in section "My questions are" how to "treat" distilled water to make it be more like tap water. For the record, I don't think distilled water. Never did.

I think a better question is why isn't it drinkable? And the answer is that it's essentially free of electrolytes. If you drink a lot of distilled water, you'll start leaching electrolytes from your body (pure dilution effect). In principle, you can just add some ions to it so that it's less of an osmotic strain on your body when you drink it.

Phys, your exact wording, except for your parenthetical, is "such that it becomes drinkable" and "such that it can become fully drinkable." Those read as health questions to me, so w/o edits to the question my close vote stands.

@hBy2Py Done. Well... You could have an easier nickname to type. =D.

Haha, I never really thought about  having to type it -- I only had to type it once. :-)

I do not see why this should be closed as a "medical" question. The OP is  refering to the actual use of desalinated seawater, ans asks how this works. (I reworded the question slighly to make this more clear.)

The content of mineral ions in tap water is usually rather small compared to the amount that you take in with your daily food. Especially calcium content varies wildly between sources, and nobody recommends calcium supplements in cities with soft, low-calcium-content water.
The common deionized (DI) water you have from a tap in a laboratory building or buy in the supermarket is not safe for drinking, firstly and mostly because it is not tested. But who would want to drink it anyway? It is more expensive, often has a taste that varies between lame and slightly foul, and you do usually have a regular drinking water tap to use instead.
The only way you could run into problems is when you have little food, drink a lot (hot environment), and have nothing to drink but distilled/rain water. On the other hand even well water likely has not enough sodium to balance what you lose through sweating.
To answer your question: A big problem with distilled water is its corrosiveness to common household tubing. Regular tap water coats everything in a layer of calcium carbonate, which makes even lead pipes relatively safe. In a standard copper pipe, DI water will first slowly dissolve the oxide layer, and then a lot of electrolytic corrosion will take place.
So, to use distilled water as tap water, you either have to make everything out of polyethylene (doesn't really work if you want hot water), or add calcium carbonate. I believe this is what's done in areas where desalinated water is fed into the communal water supply. You just have to mix in some amount of local well water.

If you were to pipe distilled water, like for a city, then the corrosiveness is a very interesting point.

Some municipal supplies (like, from glacier melt) find that adding some mineral content is a good idea, it reduces corrosion of iron pipe.

The idea that distilled water is corrosive is nonsense. And regular tap water does not contain calcium carbonate unless it is hard water; much domestic water is soft and doesn't. Oh, and soft water is often measurably more corrosive than hard water but only because it is often acidic (and even then it is only very slightly corrosive and is still entirely safe to drink).

There are several ways to make distilled water safe to drink.
(1) Pour it in a glass.
But my favorite is:
(2) Freeze the distilled water into ice cubes. Fill glass with ice cubes and add a shot of vodka to kill any remaining germs ;-) Then fill with tonic water and add a twist of lime. After you drink the mixed drink, then left the ice melt. The resulting water will be safe to drink.

While this is funny answer, I have a feeling OP was asking seriously.

Another highly scientific possibility, rather 1950ies-stlye : https://en.wikipedia.org/wiki/Irradiation#Sterilization ;-)

While working in a large industrial lab that delivered distilled water to your lab for free, we used it to make coffee - never had to de-scale the machine. And the coffee tasted pretty good - perhaps it really leached good stuff out of the coffee grounds...

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


text:
Determining the amount of water to create a given pressure in a sealed container
I have a sealed container with a volume of 1 liter. I want to go from an initial pressure of $P_{1}=1\ \mathrm{atm}$ and an initial temperature of $T_{1}=298\ \mathrm{K}$ to a final pressure of $P_{2}=9.95\ \mathrm{atm}$ and a final temperature of $T_{2}=453\ \mathrm{K}$..
I want to use water to create the pressure. But I do not know how to determine the amount of water. How would I do that?

As I understand your question, you are limited to a total volume of $\pu{1L}$ and you can only add water (which I assume should remain liquid) and heat to the system, and are able to monitor it's pressure and temperature.
Your limitation is the vapor pressure of water at the desired temperature.
At $\pu{453K}$, the vapor pressure of water is $\pu{9.88 atm}$. At $\pu{9.95 atm}$, water boils at $\pu{456K}$. So, if you heat the water to $\pu{453K}$, you will have the desired temperature and be very close to your desired pressure. Or, you could heat the water to $\pu{456K}$, and you would have the desired pressure.  Or you could heat the water to a temperature between $\pu{453K}$ and $\pu{456K}$ to get some trade off of being even closer to the desired temperature and pressure, but with neither of them being exactly what you want.
Again,this trade-off is a limitation arising from the constraint of using water to control the pressure.  If you selected a liquid with a higher boiling point at $\pu{9.95 atm}$, then you could select a temperature at which the vapor pressure of the liquid would be equal to your desired pressure.

But at 298 K the vapor pressure of water is nowhere near 1 atmosphere.

But at 373K the vapor pressure of water is 1 atmosphere, and at 456K it is 9.95 atm; I'm not sure I follow your point...have I dropped the ball somewhere?

"I want to go from an initial pressure of P1=1 atm and an initial temperature of T1=298 K"...// He can't start from there.

Yea, I guess that's a bit ambiguous, as the OP doesn't say 1 atm of what. You're right in that it can't be 1 atm of water vapor at at T=298K. If that's what he meant then question and answer are bunk, but I don't think that's what he meant.


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What solute should I use with dialysis tubing if I want a 100% retention rate?
I purchased dialysis tubing from the Carolina company.
The molecular cut off is 12,000-14,000 Daltons and I've been using table sugar for my experiments with it to raise columns of water. While the sugar does work it slowly leaks out of the membrane over time if it's left to sit for an hour or so. What should I use if I want near 100% of the solute to stay within the dialysis tubing? Preferably something cheap and non-toxic that will also dissolve in water. Or maybe a different type of membrane I could purchase that would retain the sugar better?

Polyethylene oxide, AKA polyethylene glycol or PEG has MW > 100,000. You can find higher-MW PEG in disposable diapers and similar absorbent products.
You could also use agar, pectin or gelatin, available in groceries and from lab supply companies.
One of my favorite demos was to make an agar solution with a bit of phenolphthalein added and put it in sausage casing (AKA dialysis tubing). Holding the hotdog over an open bottle of ammonium hydroxide turned it red, and over hydrochloric acid, back to a clear frank.

Thank you very much for the suggestions! Could they be used for the experiment to raise a column of water?

Sure... but for thick solutions, the rate is slow, and might need a day or two undisturbed. Also, remember that larger molecules have less osmotic effect because, for the same weight of solute, there will be fewer mols.

Then is there any membrane I could use for sugar, salt or other small molecules? I'd like it to work relatively quickly.

There are membranes capable of separating water from salts, used for reverse osmosis desalination, e.g. http://www.appliedmembranes.com/reverse-osmosis-membranes.html, and there's a guide for selecting higher-mw membranes: http://spectrumlabs.com/dialysis/dtParms.html. These may be a bit pricey compared to cellophane sausage casing, though. Try experimenting with the membrane you have before investing a lot in others.

Thank you for the resources! I found that website before but didn't realize they had roles of membrane, and the other website is great as well. Thank you for answering all of my questions


text:
Difference between isopropyl and isobutyl
Knowing the meaning of ''iso-'' prefix, i could not understand why it is used in some cases. For instance
n-propyl and isopropyl
are named according to the carbon atom which has a lack of hydrogen. The same goes for
isobutyl
where branches define the common name as well. In the end my question is: What is the logic of using iso- prefix (especially in those cases), and how can i know when to use it?
p.s- if you know the answer to my question, it is mostly appreciated to state your source of answer!

ISO is used when all carbons except last form a straight chain.

It looks like verifying my question, but could you expand this a bit? For instance what do you mean exactly with 'straight chain'?. Besides, is there any exceptions for this king of application?

You might have gotten the meaning wrong, there is no lack of hydrogen. More info in my answer.

The IUPAC Organic Chemistry Nomenclature does not explicitly define the "iso-" prefix meaning, as it is not used in the systematic naming. However, it is mentioned on several places when listing acceptable semi-systematic names. In the section A-2 (1979 Blue book, some also in 1993 Recommendations, section R-9.1, table 19a) those are (not mentioning neo- compounds)

isobutane (CH3)2CH−CH3
isopentane (CH3)2CH−CH2−CH3
isohexane (CH3)2CH−CH2−CH2−CH3

And in the section A-2.25 (some also in R-9.1, table 19b), where it lists acceptable "isoalkyl group" names:

isopropyl (CH3)2CH−
isobutyl (CH3)2CH−CH2−
isopentyl (CH3)2CH−CH2−CH2−
isohexyl (CH3)2CH−CH2−CH2−CH2−

Some related isomeric alkyls mentioned in this section are

sec-butyl CH3−CH2−CH(CH3)−
tert-butyl CH3−CH(CH3)2−
tert-pentyl CH3−CH2−CH(CH3)2−

There are also some other, unsaturated "isoalkyl" group names mentioned:

isopropenyl CH2=C(CH3)−
isopropylidene (CH3)2C=

From the above we can induce that "iso-" prefix, when related to alkanes or alkyl groups, it means the simplest (n-)alkane/alkyl isomer branched at the end of the carbon chain.
Isopropyl happens to be secondary (sec) alkyl (there's lack of carbon atoms to create primary alkyl isomeric to the n-propyl), all higher isoalkyls are primary (no branching at the alkyl "attachment").


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Is it possible to supercool hydrated salts like magnesium nitrate hexahydrate supercool just like for sodium acetate trihydrate?
Recently I stumbled upon the phenomenon of supercooling. I know that you can do this with water as well as with some phase change materials, like sodium acetate trihydrate (from the wintersport handwarmers). Reading up on some literature I saw that phase change materials are experimented with for short term storage of electrical energy from solar panels. I thought if supercooling could be used, energy could be stored for a longer period of time. But it seems that the sodium acetate trihydrate has a melting temperature that is to low and other salt hydrates with a higher melting temperature do not have the ability to supercool that easily. Arriving at my burning question to you guys is,
Why are water and sodium acetate trihydrate able to supercool and other salt hydrates are not (like magnesium nitrate hexahydrate)?
Can we increase the melting point of a salt hydrate (like sodium acetate trihydrate) without losing the ability of supercooling?

Welcome to Chemistry SE! Well, you can tell what you read or provide a link if you wish (seems you have a limit of links for now).


text:
Will cyanide substitute bromine in bromobenzene?
I've considered synthesizing benzoic acid from bromobenzene using CN-, and subsequently treating the nitrile with acid and heat. But, is this possible?

Economically, your approach is probably a bad idea.
But in general, the conversion of aryl halides to aryl nitriles does work.
I have prepared large amounts of 2,3,5,6-tetramethylterephthalonitrile from 1,2,4,5-tetramethylbenzene in a two-step procedure

iodination
reaction of the diiodide with $\ce{CuCN}$

back in the days for my PhD thesis. For the cyanation, I followed the procedure described in Synthesis, 1974, 53-55, but used N, N-dimethylformamide (DMF) as a solvent.
The halogen-cyanide exchange is know as the Rosenmund-von Braun reaction and you might find further information in a review on Cyanation of aromatic halides (Chem. Rev., 1987, 87, 779-794).

Yes, it is possible.  The Rosenmun-von Braun reaction is the classic method of doing this.  But it involves CuCN which is quite toxic.
I'd suggest that you investigate some more modern methods.  These use palladium based catalysts.  Many also use Zn(CN)2 which is still toxic but less so than CuCN.  These two are good for starters:

Iqbal, Z., Lyubimtsev, A., & Hanack, M. (2008). Synthesis of Phthalonitriles Using a Palladium Catalyst. Synlett, 2008(15), 2287–2290. https://doi.org/10.1055/s-2008-1078269
Martynov, A. G., Birin, K. P., Gorbunova, Y. G., & Tsivadzea, A. Y. (2013). Modern synthetic approaches to phthalonitriles with special emphasis on transition-metal catalyzed cyanation reactions. Macroheterocycles, 6(1), 23–32. https://doi.org/10.6060/mhc130221m

This however would be a very silly and expensive way of making benzoic acid.  Oxidation of toluene using permanganate would be much easier and quicker.


text:
Why do DMSO and acetone have such radically different melting temperatures?
DMSO and acetone have almost identical structures, except for us replacing the carbonyl carbon with sulfur to obtain DMSO. DMSO's melting point is $19~\mathrm{^\circ C}$, whereas acetone's is $-95\ \mathrm{^\circ C}$.
Is the dominant factor DMSO's larger molecular weight? Probably not. While acetone is $58\ \mathrm{g/mol}$ vs. DMSO's $78\ \mathrm{g/mol}$, 2-pentanone is $86\ \mathrm{g/mol}$, but still has a very low melting temperature of $-77\ \mathrm{^\circ C}$.
The only protons on all of these species are aliphatic, so hydrogen bonding is probably not the dominant contributor to this effect either.
What's going on?

Melting points are largerly ruled by crystal packing and thus are often contrintuitive. For example, toluene has lower melting temperature than benzene. DMSO and acetone have different molecular geometry (DMSO is trigonal pyramidal and acetone is trigonal planar) so they cannot be compared directly. DMSO also has larger dipole moment.

I seeee, okay. I totally failed to notice that DMSO's sulfur has a lone pair. That makes a more sense. Thanks!

The first thing to remember is that boiling and melting points, while seemingly simple, are incredibly hard to predict. Yes, many teachers and professors give their pupils and students a number of simple prediction rules, but there is a lot of intrinsic balance and experience associated with applying them that they may often seem unintuitive.
Comparing DMSO and acetone, the first striking difference is the replacement of carbon with sulfur. This may seem like a small change but results in radical differences. For example, compare propane with dimethyl sulfide; or the latter with dimethyl ether. $\ce{Me2S}$ is a liquid with a boiling point of around $40~\mathrm{^\circ C}$, the others are gases at room temperature and standard pressure.
There are two factors that explain the difference between $\ce{Me2S}$ and propane/methyl ether — one does better in explaining the difference to propane the other does much better in explaining the difference to dimethyl ether.

Polarity. Sulfur is more electronegative than carbon, thus the $\ce{Me2S}$ molecule is slightly polar while propane is essentially nonpolar. Since all of these molecules exhibit (approximate) $C_\mathrm{2v}$ geometry, the polarity of individual bonds does not cancel out. However, one would expect $\ce{Me2O}$ to be more polar and thus have an even higher boiling/melting point.
London interactions. These are often simplified as ‘rising with molecular weight’, but in fact, the key factor is not molecular weight (which is a nuclear property) but polarisablility of electrons. Sulfur, being larger, having an additional shell and thus being much more diffuse than both carbon and oxygen is much more polarisable. Thus, the significant difference in the boiling and melting points of $\ce{Me2S}$ and $\ce{Me2O}$ can be attributed mainly to the stronger London interactions, made possible by adding eight (core) electrons and thereby moving the valence electrons further away from the core.

Both factors play a role in the difference of melting points of DMSO and acetone, although there is even more. While acetone has a trigonal planar structure featuring a $\ce{C=O}$ double bond, DMSO is trigonal pyramidal with an additional lone pair and formal positive charge on sulfur, to give a $\ce{S^+-O^-}$ single bond. While this feature reduces the London interactions somewhat — sulfur’s higher oxidation state means that the electrons are more stable and the orbitals appear smaller at the same cutoff value — it increases polarity significantly.
Indeed, it is better to compare similar molecules; isopropanol’s structure is much closer to DMSO’s than acetone’s is.


text:
How are atoms encapsulated in buckminsterfullerene molecules?
According to nature.com, buckminsterfullerene molecules (bucky balls) can be formed by vaporizing graphite by saturating it with laser irradiation.
I haven't entirely understood why the carbon forms into a regular truncated icosahedron, but my real question is, how do scientists go about doping these "bucky balls" with molecules or atoms?

I believe the correct terminology is endofullerene, not impregnated buckyballs :) https://en.m.wikipedia.org/wiki/Endohedral_fullerene

@jjgoings, could be.  I was using a term I saw on wikipedia (not a reliable resource...)  :)  It has absolutely nothing to do with last Sunday; I wasn't around. :)  Edited.

There’s a great summary to your question in the introduction of this article:

Levitt, Malcolm H. "Spectroscopy of light-molecule endofullerenes."
  Philosophical Transactions of the Royal Society of London A:
  Mathematical, Physical and Engineering Sciences 371.1998 (2013):
  20120429.

It’s well worth a read, but I’ll summarize the basic ideas for you since I think it’s behind a paywall.
There are several techniques to get atoms and molecules in to fullerenes (or buckyballs). Note that the nomenclature for endofullerenes is A@C$_n$, where A is the atom or molecule trapped inside the fullerene and $n$ is the number of carbons making up the cage.
Early techniques involve vaporizing carbon and a metal or noble gas, and hoping that a) a fullerene would form, and b) a metal or noble gas would get trapped inside. For this reason, early endofullerenes were metallo-endofullerenes, which were fairly easy to create in low yield by laser vaporizing a metal oxide and graphite composite. This is how La@C$_{82}$ was first synthesized and reported. Fullerenes themselves are often made by a high temperature vaporization of carbon (arc discharge). If you do this in the presence of a metal, you can get metallo-endofullerenes. Some noble gases can be trapped inside a fullerene this way. They’ve actually found He@C$_{60}$ trapped in ancient rocks, which are hypothesized to have been formed before the formation of the Solar System in a similar process. These extreme techniques are limited to metals and noble gases.
Another way to get noble atoms into fullerenes involves heating fullerenes and nobles gases at high pressure. It’s thought that temporary openings in the fullerenes arise, and this allows the noble gases to insert and become trapped. You are limited to noble gases here. Yet another way is to use ion beams to “blast” atoms into the fullerenes. You can get P@C$_{60}$ and N@C$_{60}$ this way.
The above techniques give you low yield, and you are limited to metals and noble gases.
The latest techniques are often called “molecular surgery” and involve chemically creating a hole, or orifice, in a fullerene. The orifice may be enlarged by chemical means, depending on the size of the guest molecule you want to insert. These open-cage endofullerenes, which are relatively stable under mild conditions, are introduced to the guest molecules. Elevated temperature and pressure often helps get the molecules into the cage. Finally the fullerene is closed using chemical methods. In this way, H$_2$O@C$_{60}$ and H$_2$@C$_{60}$ have been made. Just in the past year, HF@C$_{60}$ was made with similar techniques.


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Arrhenius Acid-Base
According to the Arrhenius concept of acid, "an acid is a substance that,
when dissolved in water, increases the concentration of hydronium ion.
Is $\ce{HSO4-}$ in the following chemical equation an Arrhenius acid, Brønsted–Lowry acid, or both?
$$\ce{HSO4- + H2O -> SO4- + H3O+}$$
My chemistry teacher says only molecules can be Arrhenius acids and bases, is his statement true?

Second statement is false. $\ce{NaOH}$ is not a molecule but is an Arrhenius base.

So is hydrogen sulfate ion also an Arrhenius acid when reacting with water?

Yes. It has a proton to give.

Arrhenius acid-base theory is the first modern definition of acids and bases, and defines them as species that dissociate into $\ce{H+}$ and $\ce{OH-}$ ions in aqueous solution, respectively. An example of each:
$$\ce{HA ->H+(aq) + A-(aq)}$$
$$\ce{BOH(s) -> B+(aq) + OH-(aq)}$$
Brønsted–Lowry acid–base theory was proposed some time after Arrhenius acid-base theory to give a more general definition of acid-base reactions. This definition does not constrain acid-base reactions to aqueous solution. Also, it instead defines acids and bases on how they react with one another, acids defined as species that donate $\ce{H+}$, and bases defined as species capable of accepting $\ce{H+}$. The characteristic reaction:
$$\ce{HA +B <=>HB+ + A-}$$
As you can see, all Arrhenius acids will also be classified as Brønsted–Lowry acids, but there are far more many Brønsted–Lowry bases than Arrhenius bases.

Yes I fully understand both concept, the only confused pp

I fully understand both concepts, my only confusion was from my teacher's statement where he said only molecules can be Arrhenius acids and base, is the hydrogen sulfate ion an Arrhenius acid when reacting with water or not!

The term species would include molecules as well as ions.


text:
Why is bond-breaking endothermic?
When electrons gain energy they become more excited and move to a higher energy level, which increases the tendency of an atom to form a bond with another atom. Thus, surely bond making requires the absorbance of energy?
Equally when electrons lose energy they drop to a lower energy level, which would break a bond. Therefore, why isn't bond breaking exothermic?
Could someone point out the flaw in my reasoning, because I don't get why bond-breaking is possibly endothermic?

Bond making (and breaking) involve absorbance AND release of energy.  If the release exceeds the absorbance it will be exothermic and if the absorbance exceeds the release it wil be endothermic.

@JosephHirsch, Reactions may be endothermic or exothermic, but as a general rule, bond-breaking is indeed endothermic (requires energy applied to break the bonds) whereas bond-making is exothermic (outputs energy), as you said. The reaction is endothermic or exothermic depending on the difference between the total energy released by the forming of bonds and the total energy absorbed by forming bonds.

When electrons gain energy they become more excited and move to a higher energy level, which increases the tendency of an atom to form a bond with another atom.

Let me assure you that this statement is incorrect. I’m not fully sure where you extracted it from, but I assume it stems from the way many schools teach hybridisation at the beginning of organic chemistry classes; requiring an $\ce{s\bond{->}p}$ excitation in carbon from $\mathrm{[He]\ 2s^2\,2p^2}$ to $\mathrm{[He]\ 2s^1\,2p^3}$, after which the s- and p-orbitals can form $\mathrm{sp^3}$ hybrid orbitals. This idea is nothing more than a school level simplification used to get around teaching of more complex molecular orbital theory and symmetry.
Nothing prevents you from constructing e.g. a methane molecule without initial hybridisation, i.e. starting from an unhybridised carbon atom and four hydrogen atoms in a tetrahedric arrangement. I refer you to the following scheme posted in a different question and originally taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich:

As you can see on the right, carbon enters this scheme in the unhybridised ground state. There is no need to invoke a previous hybridisation before mixing orbitals; rather, it is necessary to determine the symmetry of orbitals and thereafter combine symmetry-equivalent orbitals in a bonding-antibonding fashion. Finally, fill in electrons from bottom to top.
This method will always result in stabilisation of bonding orbitals; the trade-off always being the destabilisation of antibonding orbitals in such a way that the (actual) energy gained is lower than the (virtual) energy lost.
Therefore, assuming a positive bond order, forming a bond typically liberates energy while breaking one will typically require energy. I am not aware of any counterexamples, but the sentence is phrased so that it remains true when the obligatory counterexample is posted as a comment.

Sorry, your logic doesn't quite hold. Increasing the energy of electrons does make it more likely that they will do something, but the end result is frequently something that is lower in energy than the starting state.
Dropping in energy levels doesn't break a bond. Generally, bonds correspond to the lowest energy levels.
The main thing to keep in mind is that one way to define a bond is that it is the stabilization of electrons between a group of atoms. For neutral species, the stabilization is relative to the energies of the electrons in the atomic orbitals of the constituent atoms.

No matter how weak the bond is, there are always going to be some interactions between the two species involved in the bond. It is because of those interactions that the bonds were formed at the first place so energy will always be required to break those interactions and  hence bond cleavage is always endothermic.

You seem to have it the wrong way round. Adding energy to an electron in a bond places it into an anti-bonding orbital making it more likely that the bond will break. Adding a second amount of energy to put two electrons into anti-bonding orbitals even more so. The normal bonded state is the lowest energy state by convention this is the most negative energy. When a bond is formed energy is released and is normally taken up by the surrounding molecules and translational, vibrational and rotational energy.

To understand this, first you have to know that energy of a system is always inversely proportional to the stability of system.
When two atoms come towards each other, the energy of the system of two atoms decreases ( Here the energy is potential energy). When atoms form bond, this energy becomes minimum ( since the system of atoms is most stable now). Notice that energy has become minimum so there must be some lost of energy, and this is the energy which is released when a bond is formed.
Now if you want to to break that bond , you have to seprate those atoms ( separating atoms means decreasing the stability of two atom system) and since stability is inversaly proportional to energy hence decreasing the stability is equivalent to increasing the energy. So when you are breaking a bond, you are seprating the atoms and this will lead to an increase in energy.  An increase in energy of system is only possible if energy is provided to system. I shall let you conclude now.

You have to break the mutual electrostatic attraction between the electrons and protons of each atom.

A bond between two atoms happens because they get a more stabilized stadium of energy. In chemistry, low potential energy means more stabilization. Think to a ball at the bottom of a basin. This is an extremely stabilized system, and means you will have to put energy to move this ball.
Conversely, a ball on top of a potential energy hill is not a stable system, and no energy is required move the ball down.
A bond between two atoms is the ball at the bottom of a basin - energetically speaking.


text:
UV-Transparent Plastic?
I am looking for a type of plastic, preferably in the form of a sheet or film, that is weatherproof, durable, flexible, and both transparent and UV-transparent. Most films I've seen so far are treated to be UV-resistant, where can I find untreated plastics for purchase?
Planning to use this plastic in order to cover a solar panel, without inhibiting energy output.

The number of UV photons compared to those in the visible is very small, not only because of the black-body distribution from the sun but also because atmospheric ozone absorbs a lot of UV. So it may not be a cost effective solution since PMMA sheet is expensive. (They are treated to be UV resistant to extend their lifetime against UV induced photochemistry)

Did you consider PMMA?
A 3 mm Plexiglas G UVT layer seems to have a transmittance > 80% at $\lambda = 300~\mathrm{nm}$ and above.

I concur with Klaus's recommendation, PMMA is likely your best bet.
The UV disposable cuvettes we buy here are from BrandTech; below is a UV transmission chart for various materials from one of their marketing brochures (click image to enlarge):

They've clearly had success in developing materials with even better UV transmittance (green and yellow curves) than PMMA (red curve), but it looks like the latter may be your best option.


text:
What gas are the bubbles that evolve in the redox reaction of copper(II) chloride and aluminum?
When aluminum (foil) is placed in an aqueous solution of copper(II) chloride, the aluminum atoms lose electrons to the copper(II) ions and replace them to form aluminum chloride, and to free up metallic copper.
The metallic copper has a tendancy to float on the surface of the solution, being buoyed by gas bubbles. The bubbles appear to evolve from the interface of the aluminum foil with the solution. What is the source and nature of the gas bubbles?
I have two hypotheses:

The bubbles are coming from dissolved $\ce{CO2}$ that is coming out of solution due to an increase in temperature close to the site of reactions.

The bubbles are $\ce{O2}$ bubbles coming from the layer of aluminum oxide on the surface of the aluminum foil.

If this is not the case, what happens to the portion of the portion of the foil that was in oxide form prior to the reaction.  How would the aluminum oxide react with copper(II) chloride?

See http://chemistry.stackexchange.com/questions/68745/redox-mgcopperii.  Copper salts give weakly acidic solutions from which strongly electropositive metals can displace hydrogen.

I think the gas is hydrogen(H2).

Do not think but think it out and give a chemical reason.

The currently accepted answer does not consider the possibility discussed here, where magnesium is the metal and again there are bubbles.  Transition metal salts of strong acids are weakly acidic in water, and that weak acid enables a strongly electropositive metal which reacts only slowly or not at all with pure water to begin displacing hydrogen more readily. In the case of copper(II) chloride tetrahydrate a 50 mg/L solution has a pH between 3 and 4 according to supplier material data.  Thus the aluminum in this case is displacing hydrogen (from the acid produced by hydrolysis) as well as copper.

They do not appear to be flammable.  There is no insoluble Aluminum oxide or hydroxide at the end of the process.  Could you show me what all of the predicted products would be based on your hypothesis?

Aluminum --> hydroxo complexes , $\ce{Al(OH)_2^+}$ etc.  Magnesium --> cation stays in solution below pH 8 or 9.

Some people are never satisfied. I am one of those people when I see a drive-by downvote with no suggested alternative or where the answer needs to be improved.

You stated an observation with no backup. Read my answer below. These reactions will happen between steel and stainless steel with even slight conductivity. The steel corrodes. The hydrogen is literally absorbed into the nickel containing SS. Agreed acid hastens the process but the catalytic overvoltage lowering hastens it more.

Then I am returning the favor. Where isthe backup for your claim? In another answer you comment that aluminum does not form an amalgam, which is false. Good day!

Metallic aluminium reacts with water somewhat like sodium $\ce{Na}$, according to $$\ce{2Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$ So it produces a lot of gaseous hydrogen $\ce{H2}$.
Of course, usually, aluminum metallic pieces do not react with water. This is due to a thin, waterproof and continuous layer of aluminium oxide $\ce{Al2O3}$ which is automatically produced on aluminium metal by the atmospheric oxygen. It prevents the contact metal-water. This oxide layer can be removed by dipping the aluminium piece into a $\ce{HgCl2}$ solution. It makes an $\ce{Al}$ amalgam reacting quickly with $\ce{H2O}$ and produces long white filaments of $\ce{Al(OH)3}$ growing at a quick glance, as long as there is some metallic mercury remaining on the aluminium surface.

You almost got it but did not consider why the mercury remains in contact. If Na or K amalgams are representative amalgamation lowers activity and Al supposedly does not form an amalgam.

The gas bubbles are very likely just air that was dissolved in the water previously and now gets released. I would expect most of the bubbles to actually be nitrogen, followed by oxygen, argon and then some carbon dioxide — that is assuming you used deionised water which should not have significant amounts of carbonate and hydrogencarbonate dissolved.
The copper particles merely supply suitable nucleation sites for the gas.
The theory on aluminium oxide being oxidised to pure oxygen is not correct. Oxygen is a very oxidative element meaning that you require strong oxidants to oxidise oxide anions back to oxygen. With ‘strong’, we’re talking the oxidative strength of potassium permanganate. Aluminium’s oxidative powers are nowhere near that strong.
Copper(II) chloride and aluminium oxide do not react with each other; only metallic aluminium beneath the passivisation layer is able to react with copper(II) ions.

I was thinking that given the temperature it could also be water vapor.

@JosephHirsch Water vapour does not form bubbles in water at room temperature. You didn’t mention that you were boiling your solution.

Not boiling.  The reaction is exothermic.  Not violently but I suspect that it gets warm enough at the reaction sites.

@JosephHirsch No.

I object. This is most probably hydrogen.

I will do more than object.  I am directly going to post my own answer and refer to my answer with mangnesium.  That answer currently is higher scoring than this one.

An active metal such as iron, zinc, Aluminum, magnesium reacts slowly with water to form H2 and the metal ion or oxide-hydroxide. The rates and products depend on the hydrogen ion concentration. There is an over voltage to this reaction that is reduced if a more noble metal is in contact with the active metal. This is how sacrificial anodes work; the protected metal is more noble. Commonly Zn, Al or Mg are used to protect the more noble Fe and the production of H2 is ignored.  Typical noble metals are iron, stainless steels, nickel, copper, the palladium and platinum groups; for some reason gold and silver do not work as well. Any situation where different metals are in contact, with an electrolyte present, will result in corrosion and usually the reduction of hydrogen ion to give H2 gas on the more noble metal.
In the reaction of Cu++ ions with Al metal some copper immersion-plates on the aluminum. This combination catalyzes the reaction of the aluminum metal with the aqueous solution, generating H2. There are two competing reactions.


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What elements (in the pure form) would harm a human, if eaten?
What pure elements, if we eat it in a relatively small dose (around a piece of sugar), can be harmful/lethal for an average human? Is, for example, eating pure carbon bad for the organism?

Related: Are there any good examples of commonly ingested molecules that contain particular toxic individual elements?

I think you need to be more specific with the question. Lots of things will kill you or could cause harm. Eating a diamond heated to 3000 K is probably bad. If you eat a large mass of carbon-14, that's probably also bad.

Is, for example, eating pure carbon bad for the organism?

No, quite the contrary.
Activated charcoal (in the form of powder or tablets) was/is traditionally used to treat diarrea. It is listed on the WHO Model List of Essential Medicines on page 4.

Another fun data point: polonium-210 has $\mathrm{LD}_{50}$ of approximately $50\ \mathrm{ng}$ by ingestion or $10\ \mathrm{ng}$ by inhalation. Link
Obviously, the answer is yes for some elements.

I hear its all about the dose, even water can kill you if you drink too much.
And cyanide wont kill you if its a small dose, https://www.youtube.com/watch?v=bWNpO5vvhpk

OP is asking about small dose

Here are some more elements that probably would cause no harm if you ate a sugar-lump sized piece: B, Mg, Al, Si, S, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Pd, Ag, Sn, Pt, Au, Bi. Probably there are many more. Don't know about the rare earth elements. Also, all the noble gas elements, plus hydrogen, are nontoxic.

Sugar-lump sized pieces of most metals would probably break your teeth or get stuck in your throat. Other than that, those you listed are harmless indeed.

Copper isn't entirely benign; there is some literature on human toxicity (and, of course, it kills plants: copper sulfate is sometimes used to keep roots out of sewer pipe).   Trace amounts are beneficial, though.

Mg may give you gas (reacts readily with gastric HCl) and too much may displace calcium in bones and teeth.

5 grams elemental at STP assuming no physical trauma:
Immediate  serious harm: P and group 1 metals and Calcium.

It would be good to mention that you mean white phosphorus. Other forms are relatively harmless.

Regarding different allotropes, He would be harmless, but I would avoid eating 3 grams of $\ce{He2}$.

Unstable particles are usually not counted as allotropes.


text:
How does electrophoretic separation work in terms of separating soil into minerals?
I found this diagram and would like to know how it works with soil.
Google searches yield protein/biology things, anyone know how this method would work with soil?

The search term: "continuous particle electrophoresis" yielded much more results.

electrophoresis - the movement of charged particles in a fluid or gel under the
influence of an electric field.

source: http://www.youtube.com/watch?v=LNUgcq1M4LI
-

source: http://www.nature.com/articles/srep19911/figures/1
Heard some guy invented it while messing with some clay, but couldint find any video examples of that.


text:
Why are alkali salts so soluble?
There's a certain solubility rule stating all carbonates, phosphates and hydroxides are insoluble unless mixed with hydroxides or alkali metals.
Why is this? What makes the alkali metals so special?

So the reason for carbonates and phosphates being insoluble in water because in an acidic environment, they are both protonated and lose their negative charges. The symmetry of their binding groups produces identical resonance structures at high pH values (since rotating one double bond will shape it into a superimposable structure of another). This allows any given oxygen at a time to posses the double bond (in reality the pi bond to the phosphorus or carbon is shared throughout all the bonded oxygens and each oxygen maintains an equal distance away from the central atom).
This partial negative charge (created by lack of firm placement of the pi electrons) can be neutralized by bonded hydrogens (this will create a small positive charge, however hydrogen's ionization energy is much less than oxygen's electron affinity). The lack of polar-ness of the phosphates and carbonates makes them insoluble in water.
In a basic environment, (created by hydroxides) the carbonates and phosphates will become deprotonated and charged. The charge on these molecules will allow them to be more miscible with water.

https://en.wikipedia.org/wiki/Carbonate

Since purely ionic bonding cannot exist, the percent ionic character of a bond is often useful when comparing the salts of different metals. For reference, an electronegativity difference of 1.7 is estimated to have about 50% ionic character, and 50% covalent character$^{[1]}$.
This being said, alkali metals comprise the least electronegative group in the entire periodic table (see here). This gives these metals some of the greatest electronegativity differences when forming bonds. The electron pair is so closely attracted to the electron acceptor that they are practically transferred to it.
The general trend is for electronegativities to decrease down the group, making lithium the least electronegative of the alkali metals. As such, some of the salts of lithium have different properties: $\ce{LiF}$ is not soluble in water, $\ce{LiI}$ is soluble in most organic solvents, and $\ce{LiOH}$ is the only non-deliquescent alkali hydroxide$^{[2]}$.

${[1]}$ Wikipedia, Ionic Bonding, Comparison with Covalent Bonding
$\ce{[2]}$ Wikipedia, Alkali Metal, Electronegativity


text:
is avogadro's law for the same gas?
Avogadro's law is V1/n1 = v2/n2 - volume and number of  moles are directly proportional. My question is - can this law apply to two different gases - for example, Hydrogen with volume v1 and number n1, and Oxygen with volume v2 and number n2 - are they proportional?

A quick google search answers this. From Encyclopaedia Britannica: Avogadro’s law, a statement that under the same conditions of temperature and pressure, equal volumes of different gases contain an equal number of molecules.

Since Avogadro's Law is speaking about ideal gases then it's true for all gases that behave ideal.
Since ideal gases don't interact with each other and don't have any volume for single molecules (they only have a volume as a whole) Avogadro's Law holds for different gases.
Since for real gases the approximations of the ideal gas don't hold then Avogadro's Law doesn't hold. Yet a ratio similar can be achieved by dividing V over n of different gases to get the molar Volume and by using different equtions to describe real gases a ratio factor can be achieved for a given pressure and temperature.

Avogadro's Law says just that.  Regardless of the gas equal volumes have equal n.  (At constant p and t) and v/n is constant.
v/n=k will be true for different amounts of the same gas and approximately true between two gases.


text:
How to find the temperature change of a solution?
Right now I'm doing a calorimetry lab and for the pre-lab we have been asked to determine the temperature change (in Celsius) of ammonium chloride in water.
The question asks for the expected temperature change ($\Delta T$) of $\pu{8.5 g}$ of $\ce{NH4Cl}$ in $\pu{100 mL}$ (or $\mathrm{g}$) of water, with the molar enthalpy ($\Delta H_\mathrm{sol}$) of the solution being $\pu{0.277 kJ/g}$. The conversion of $\mathrm{kJ/g}$ is throwing me off and I cannot figure out how to solve for $\Delta T$ with the given info.
I know that there are $0.165048...$ moles of solution, which gives me everything I need to solve. We were given the equation
$$n \Delta H_\mathrm{sol} = mC \Delta T,$$
where $m$ - mass of water and $C$ - specific heat capacity of water. I am assuming it needs to be rearranged to
$$\Delta T = \frac{mC}{n\Delta H_\mathrm{sol}}$$
Any help is greatly appreciated and I can explain in further detail if necessary. Sorry for the Celsius, we apparently don't use Kelvin in our calculations.

There is no such thing as mole of solution.

@IvanNeretin Sure there is. If I have a mixture of chemicals that sum to 6.022 x 10^23 molecules, then I have one mole of solution.

The main problem here is a simple mistake in the algebra. You rearranged:
$\pu{n\timesΔH_{sol} = m\times C\timesΔT}$
to
$\pu{ΔT = \frac{m\times C}{n\timesΔH_{sol}}}$
rather than
$\pu{ΔT = \frac{n\timesΔH_{sol}}{m \times C}}$
Additionally, the molar enthalpy of solvation is given in the problem as units of kJ/g, not kJ/mol. Presumably, this is an error in the given problem. According to Parker, V . B ., Thermal Properties of Uni-Univalent Electrolytes, Natl .
Stand . Ref . Data Series — Natl . Bur. Stand .(U.S.), No .2, 1965, the molar enthalpy of solution for $\ce{NH4Cl}$ is $\pu{14.78 kJ/mol}$.
There was also a slight miscalculation in the moles of solute. Where you calculated $\pu{0.165 moles}$ of $\ce{NH4Cl}$, you should have gotten:
$\mathrm{8.5 g / 53.49\frac{g}{mol}=0.159~mol}$
These errors corrected, plugging in the values to solve for $\Delta \text{T}$ is trivial and gives:
$\pu{ΔT = \frac{\pu{0.159 mol}\times \pu{14.78 kJ mol-1}}{\pu{100g} \times \pu{4.186J~g-1~K-1}} = 5.6K}$

I agree with Airhuff (almost) completely.
It doesn't really matter if you use kJ/mol, or kJ/g. As long as your units can cancel out.
$$\frac{14.78kJ/mol}{53.491g/mol} = 0.277kJ/g$$
But I mostly wanted to point out, the reason why it is okay to use Celsius for this calculation - because you have "ΔT" in your expression.
Say you have something at 30°C, and it changes to 24°C.
ΔT = 24°C - 30°C = -6°C
In Kelvin, the temperatures are 303K or 297K.
ΔT = 297K - 303K = -6K
The way I would do it:
$$ΔT =8.5g NH4Cl * \frac{0.277kJ}{1gNH4Cl} * \frac{1}{100g H2O}  *\frac{1gH2O * °C}{0.004184kJ} = 5.627°C$$


text:
Naming of ester alkoxide
I'm trying to name the following compound:

I believe the carboxyl group has priority and should make up the parent name. I'm not sure how to name the alcohol derivative portion. My first thought is:

potassium ethyl 2-oxyacetate

or

potassium ethyl oxyacetate

I've seen that there is a compound without the alcohol derivative and ester side chain called glycolic acid, but I have seen no examples of this common name being used as part of an alkoxide salt.
What is the correct way of naming a compound like this?

As you have already mentioned, the key to the answer of this question is the seniority order for classes that are expressed by suffixes.
Esters are senior to hydroxy compounds (including alcohols). Therefore, the preferred IUPAC name for the corresponding glycolic acid ethyl ester with an intact hydroxy group is ethyl hydroxyacetate. (Note that according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the locant ‘2’ in ‘ethyl 2-hydroxyacetate’ is not cited.)

However, anions are senior to esters. Therefore, the anion formed by subtracting a $\ce{H+}$ from the hydroxy group of ethyl hydroxyacetate is named as ethanolate. The ethanolate is substituted using the usual principles of substitutive nomenclature. The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix. The group $\ce{-O-CH2-CH3}$ is indicated by the prefix ‘ethoxy’. This leads to the systematic name 2-ethoxy-2-oxoethan-1-olate for the anion and potassium 2-ethoxy-2-oxoethan-1-olate for the salt.


text:
With molecular formula, how do I know if it linear or one element surrounded by other elements?
If I was given a molecular formula (i.e, PoOBr2), how do I know if the Lewis structure is linear (4 in-a-row, basically a straight line) OR it is an element surrounded by other 3 elements?
I know the answer for this is Po in the central and surrounded by other elements.

The use of polonium in an example like this is quite unusual. I'm not even sure this compound is molecular since polonium is quite metallic. Originally, I was thinking phosphorous, but that would almost certainly be $\ce{POBr3}$.

@Knox777 How will it be linear?

I have to admit, I am not sure how Po gets to 4 bonds, but I know I've seen it.  I think it forms $PoBr_4$ and also nothing else works here.

The formula is insufficient. Consider pentane, $\ce{C5H12}$, which occurs in straight and branched isomers, the extreme case being neopentane, in which the central carbon is linked only to four methyl groups.

For formulas that can produce only one isomer or "connective pattern" and  using the typical Lewis structure bonding for the elements, start by separating the atoms into those that only make 1 bond and those that make more than one bond.  In this case:
Elements that make more than one bond have to be bonded together with at least one bond, so in this case take the Po and O and make one bond between them while leaving the other open bonding sites alone like this:
Now you have to place the monovalent elements on the multivalent elements.  You could do this:
but you have used up all of your atoms, and Polonium still needs to make 2 bonds, and it can't do it by itself, so you can see that IF you put them all in a row, you would get stuck with an atom that has to make 2 more bonds but has nothing to bond to. Instead, we need to move both of the Brs to the Po like this:

Then since Po and O each need to make one additional bond, they can form a second bond between them like this:

You can turn around the Br symbols and also make the bonding symmetrical if you want to:
(Get in the habit of showing your unbonded electron pairs at least for atoms that make more than 1 bond like Oxygen and Nitrogen)


text:
Does temperature affect heat capacity?
My chemistry  book says that heat capacity varies with temperature. There is no further discussion.
Does the heat capacity of a given amount of substance really vary with temperature? If yes, why is it so?

Because, well, why not? Many properties of substances vary with temperature; so does heat capacity. On a deeper level, as you rise the temperature, more and more degrees of freedom unfreeze, so the heat capacity grows. (Water vs vapor is a notable exception.) At zero K, heat capacity of anything tends to 0.

Related: Can we define a new unit instead of specific heat that is constant with temperature rise?

The heat capacity is the slope of the plot of internal energy U with temperature T. The internal energy is energy that due to the rotation and vibrational energy a molecule possesses and as the temperature increases more rotational and vibrational energy levels becomes excited and so the internal energy increases.
Initially the internal energy is small and changes slowly as energy is added, mainly because there is not enough energy to excite many rotations or vibrations. Thus the slope of U vs. T is small and so is the heat capacity.  Then as temperature increases more of the energy levels become excited & the internal energy rises rapidly and so does the slope of U vs T and so heat capacity increases.
At very high temperatures  many, many energy levels are populated and so the rate of increase becomes constant as the temperature increases and so the heat capacity becomes constant, i.e. it levels out to a maximum value.  The plot shows the change of heat capacity with energy just for vibrational levels, but it has the same general shape when rotational and translational energy are added.

internal energy vs temperature

vibrational heat capacity vs temperature

low temperature region of heat capacity vs temperature.The 'bumpy' plot is due to the discrete values of vibrational energy levels


text:
Band gap difference in different allotropes of Group4 materials
Diamond and bulk Si are both Group 4 elements found in their $sp^3$ hybridized states, having bandgaps of around 5.3eV and 1.1eV respectively, but Graphene and Silicene are both $sp^2$ hybridized allotropes of the said elements and are zero bandgap materials. What promotes this change in behaviour? Is it the change in hybridization? or the super-aromaticity??

Related: https://en.m.wikipedia.org/wiki/Borophene

@QuantumAMERICCINO It doesn't say what promotes this behaviour.


text:
Enthalpy Change Of Formation
Enthalpy change of formation means the heat change when 1 mol of the compound is formed from its elements under standard condition.
For example, the enthalpy change for the formation of carbon dioxide gas  is -393.5 KJ/mol
But, I'm confused as the enthalpy change of any element under the standard condition is zero, so shouldn't the enthalpy change of formation of carbon dioxide is also zero as well?
I am missing something but I just couldn't figure out what it is

You know what's an element, right?

$$\ce{C(graphite) + O2(g) -> CO2(g)}$$
$$\Delta H^{\varnothing} = -393.5\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$
That is, the reaction is exothermic.
Since the reactants are elements in standard state, the heat of formation for carbon dioxide is the same as the enthalpy change of the above reaction.


text:
Time dependence of integrated rate equations
Is it possible for an integrated rate equation to not be dependent on time? What law or principle says so in chemical kinetics?
What if an integrated rate equation obtained only depends on the concentrations of the reactants? Is it possible? Can you refer me to books and references?
By definition, an integrated rate equation shows the change in concentration as a function of the progress of the reaction (time). So I don't understand why professor insists that an integrated rate equation can exist without having the variable time in it.

If that's really what the Prof. said then it doesn't make any sense. The only special case which is usually covered is that the half-life of a first order reaction is not dependend on concentration. But that's a totally different statement so I don't think that's what was said/he wanted to say.

Hello, if you look at the derivation posted by the thread starter at http://chemistry.stackexchange.com/questions/62495/determination-of-the-integrated-rate-equation-for-a-reaction/62584?noredirect=1#comment107680_62584 , you will see that at the end, he obtained an equation independent of time and so, he had an integrated rate equation which is not a function of time. However, I think that this is fundamentally wrong but I don't have a strong argument or a reference that explicitly tells that this is wrong. That's why I cannot defend my stand.

Under steady state conditions it is assumed that $d[C]/dt =0$ so time is removed this way. However, this only applies under certain conditions and only for some reactions. Secondly, if an equilibrium is involved, at long times when equilibrium is established concentrations are effectively constant.  Other than these cases the concentration of all species vary with time.

Hey porphyin. Yeah by that condition the concentration of C does not depend on time, but the reactants A and B still do. I think you agree that the concentration of reactants are still time dependent and thus, an integrated rate equation without time variable (except when already in equilibrium) is problematic

From a mathematical standpoint: If the integrated rate law is independent of time, then the rate of change of the concentration must equal zero (by the fundamental theorem of calculus). So you're looking for a reaction in which the concentration of the species you want in the rate law (presumably the reactant) does not change appreciably.
As others pointed out, this is a common case for intermediates under the steady-state approximation (rate of production = rate of consumption, net rate zero). But that's just for intermediates. For a substrate, you would need the substrate concentration to be essentially constant. That's probably not a "reaction" that will be interesting to study. A trivial solution, if you will.

Ok, so if we got the reaction
$A\rightleftharpoons B$
with $k_1$ for $A \rightarrow B$ and with $k_2$ for $B \rightarrow A$ then
$\frac{d[A]}{dt}=-k_1 [A]+ k_2 [B]$
if you integrate that you end up with
$k_1 [A]=k_2 [B]$
which is time independent

This is making the assumption that the initial concentration of [A] is the equilibrium concentration. If the system is initially not at equilibrium then [A] varies with time: http://i.stack.imgur.com/L7X6c.jpg

yes ofc. I wanted to write that that's the case after the equilibrium is reached, somehow forgot that.

So is equilibrium only the case wherein reaction rate is time independent? Since effectively the change in concentration of all reacting species per unit time is zero, right?

@DSVA I haave a trivial question about integrated rate laws...so trivial that I searched for "integrated rate laws" and found this question relevant, so to comment. why is the rate law integrated than to differentiated? R is the instantaneous rate of concentration function against Time. R is a derivative. why is it integrated then? what is the intention?