?

Personally, I think that a bespoke pair of seamless wholecut plain toe good year welted leather sole oxford (with 3-5 eyelets) in black or dark brown is the most formal/elegant shoe you can ever own. I'll list some reasons below why I think that and also, here's an example to help picture what I'm talking about:

http://www.theshoesnobblog.com/2014/08/t...

Let me know what you guys think!

1) Bespoke because it's made to perfectly fit your feet, unlike ready to wear shoes

2) Seamless Wholecut: Wholecut means that only one piece of leather was used, so it has to be free from any imperfections. Combined with the fact that it's seamless means that one piece of leather must be larger than usual so that there's no stitch on the back heel part of the shoe. No machine can make a seamless back heel. Only a person's hands can make a seamless back heel, thus making it strictly bespoke

3) Plain Toe because less decorations means more elegance, combined with the fact it's already a seamless wholecut

4) Good year welted to make it more water proof and easier to resole at a lower cost over the life of the shoe

5) Leather sole because it's more durable and less chunky than a rubber sole

6) Oxford because of its closed lacing system. Now, I don't know if the number of eyelets affects formality but if it does, let me know please

7) Black would be the most formal because it's the darkest color but Dark Brown would be nice too

Personally, I think that a bespoke pair of seamless wholecut plain toe good year welted leather sole oxford (with 3-5 eyelets) in black or dark brown is the most formal/elegant shoe you can ever own. I'll list some reasons below why I think that and also, here's an example to help picture what I'm talking about:

http://www.theshoesnobblog.com/2014/08/the-seamles...

Let me know what you guys think!

1) Bespoke because it's made to perfectly fit your feet, unlike ready to wear shoes

2) Seamless Wholecut: Wholecut means that only one piece of leather was used, so it has to be free from any imperfections. Combined with the fact that it's seamless means that one piece of leather must be larger than usual so that there's no stitch on the back heel part of the shoe. No machine can make a seamless back heel. Only a person's hands can make a seamless back heel, thus making it strictly bespoke

3) Plain Toe because less decorations means more elegance, combined with the fact it's already a seamless wholecut

4) Good year welted to make it more water proof and easier to resole at a lower cost over the life of the shoe

5) Leather sole because it's more durable and less chunky than a rubber sole

6) Oxford because of its closed lacing system. Now, I don't know if the number of eyelets affects formality but if it does, let me know please

7) Black would be the most formal because it's the darkest color but Dark Brown would be nice too

I know that all moving objects have linear velocity, whether moving in a straight (just linear velocity) or curved line (both linear and angular velocities). I'm just not sure what a detailed explanation of it is. So I'd like to give my take on it and a few questions.

To my understanding, linear velocity is straight line (no directional change) or 1-dimensional motion and it can consist of either linear "average" velocity or linear "instantaneous" velocity. So if an object travels north in a straight line and then east in a straight line (thus 2 straight lines), would the velocity of the object still be considered linear despite the directional change? Also, is this north-east motion still considered 1-dimensional motion or is it now 2-dimensional?

Also, people usually say that the magnitude of the velocity is equal to the speed but that's not always true. The magnitude of "average" velocity is not the "average" speed, except when the object moves in 1 straight line with no directional change (thus when there's "linear" velocity?). And as always, the magnitude of "instantaneous" velocity is always equal to the "instantaneous" speed.

To my understanding, I there are 2 types of velocities:

1) Linear Velocity (which involves straight line/1 dimensional motion): This in turn can be broken down into average and instantaneous linear velocities.

2) Angular Velocity (which involves curved motion): This can also be broken down into average and instantaneous angular velocities.

And I know the two are related via the equation, v = ωr

*Is my breakdown correct?

**Also, if linear velocity means "straight line/1 dimensional" motion, this also means the object is not changing direction. However, couldn't an object go "left" in a straight line and then "up" in a straight line and have this right-up motion still be considered linear velocity despite the directional change? Or does the object's motion have to be solely straight left or solely straight up in this case? If it's the former, wouldn't the motion then be 2-dimensional as opposed to 1-dimensional?

***For angular velocity, does the motion have to be "circular" or can it be any type of curvature?

****Everywhere I look, I keep seeing that "the magnitude of velocity is speed." But I think people are getting sloppy b/c shouldn't it be that the magnitude of "instantaneous" velocity is ALWAYS instantaneous speed while the magnitude of "average" velocity is NEVER average speed with the exception of when an object moves in one straight line (and thus "linear" velocity??), b/c in that case, the magnitude of displacement and distance will be the same.

So let's say our slow elementary step is:

aA ---> bB (slow)

So our average Rate for our reactant A will be -ΔA/aΔt and if we just take the limit as Δt goes to zero we get the instantaneous Rate being -dA/a dt

Also, using our slow step, we can also say the instantaneous Rate = kA^a

So setting them equal we get:

-dA/a dt = kA^a

Now my question is, when our coefficient "a" is "2" (as in a bimolecular slow elementary step), we would have -dA/2dt = kA^2, which would integrate to 1/At - 1/A0 = 2kt.

However, the actual second order integrated rate law is 1/At - 1/A0 = kt. Why do all derivations exclude the 2?

**Also, how would a slow elementary step for a zero order reaction look like? (is it 0A ---> B? if so, it wouldn't make sense to make the average rate as -ΔA/0Δt b/c zero in the denominator makes the rate undefined)

So let's say our slow elementary step is:

aA ---> bB (slow)

So our average Rate for our reactant A will be -ΔA/aΔt and if we just take the limit as Δt goes to zero we get the instantaneous Rate being -dA/a dt

Also, using our slow step, we can also say the instantaneous Rate = kA^a

So setting them equal we get:

-dA/a dt = kA^a

Now my question is, when our coefficient "a" is "2" (as in a bimolecular slow elementary step), we would have -dA/2dt = kA^2, which would integrate to 1/At - 1/A0 = 2kt.

However, the actual second order integrated rate law is 1/At - 1/A0 = kt. Why do all derivations exclude the 2?

**Also, how would a slow elementary step for a zero order reaction look like? (is it 0A ---> B? if so, it wouldn't make sense to make the average rate as -ΔA/0Δt b/c zero in the denominator makes the rate undefined)

So adding equal volumes of ethanol and water results in a solution with a volume less than the mathematical volume. I understand that it has to do with intermolecular forces, mainly Hydrogen-forces, and how they form in the solution. However, there's Hydrogen-forces within the pure liquids themselves prior to mixing. A molecule of H2O can form a maximum of 4 Hydrogen-forces (2 H-force donors and 2 H-force acceptors) with (obviously) 4 water molecules, while a molecule of C2H5OH (ethanol) can form a maximum of 3 Hydrogen-forces (1 H-force donor and 2 H-force acceptors) with 3 neighboring ethanol molecule. So when we mix the two equal volumes of solution, we still have Hydrogen-forces. How do these Hydrogen-forces in solution lead to the molecules of water and ethanol becoming more efficiently packed and thus take up less volume than by themselves? And since there's a smaller volume after mixing, does this mean the solution is "more" dense/How would the density of the solution be relative to each liquid?

I know that water molecules (18 g/mol) are smaller than ethanol (or any alcohol molecule) molecules (46 g/mol) and that water has a density of 1 g/ml while ethanol has a density of 0.789 g/ml.

So adding equal volumes of ethanol and water results in a solution with a volume less than the mathematical volume. I understand that it has to do with intermolecular forces, mainly Hydrogen-forces, and how they form in the solution. However, there's Hydrogen-forces within the pure liquids themselves prior to mixing. A molecule of H2O can form a maximum of 4 Hydrogen-forces (2 H-force donors and 2 H-force acceptors) with (obviously) 4 water molecules, while a molecule of C2H5OH (ethanol) can form a maximum of 3 Hydrogen-forces (1 H-force donor and 2 H-force acceptors) with 3 neighboring ethanol molecule. So when we mix the two equal volumes of solution, we still have Hydrogen-forces. How do these Hydrogen-forces in solution lead to the molecules of water and ethanol becoming more efficiently packed and thus take up less volume than by themselves? And since there's a smaller volume after mixing, does this mean the solution is "more" dense/How would the density of the solution be relative to each liquid?

I know that water molecules (18 g/mol) are smaller than ethanol (or any alcohol molecule) molecules (46 g/mol) and that water has a density of 1 g/ml while ethanol has a density of 0.789 g/ml.

Carbon has an electron configuration of (1s2) (2s2) (2px1 2py1 2pz0)

Thus, only two (unpaired) valence electrons total (from 2px and 2py) are available for bonding. However, in order for Benzene, C6H6, to form, each carbon must have three unpaired electrons ready to bond with two other carbons and one hydrogen. Thus, the 2s orbital (initially with paired electrons) of each carbon will have one of its electrons excited and placed into its empty 2pz orbital, resulting in the 2s, 2px, 2py, & 2pz pure atomic orbitals of each carbon containing one (thus unpaired) electron each. Since each carbon will covalenty (sigma) bond with two other carbons and one hydrogen, only "three" pure atomic orbitals per carbon will be needed for hybridization; the 2s, (arbitrarily) 2px, and 2py. After constructive and destructive interference of the selected atomic orbitals occurs, "three" sp2 hybrid/unpure orbitals per carbon will be yielded and will orient themselves (due to electron repulsion derived from VSEPR theory) to the apexes of a trigonal planar shape (120 degrees apart). Now each carbon can use two of it's sp2 hybrid orbitals to overlap DIRECTLY with two adjacent carbons and the remaining sp2 hybrid orbital to overlap DIRECTLY with a 1s atomic orbital of a hydrogen to form SIGMA covalent bonds, yielding the hexagonal ring. The remaining 2pz pure atomic orbitals (of each carbon) with their unpaired electron will be situated perpendicular to the trigonal planar geometry of each of the six carbons that are sp2 hybridized. These six "in-phase" 2pz pure ATOMIC orbitals (which are parallel in relation to each other) of each carbon will LATERALLY/side to side overlap and form Pi covalent bonds that yield six MOLECULAR ORBITALS that are delocalized above and below the plane of sigma bond framework holding the sp2 carbons in the ring.

**What I don't understand is why the sp2 hybrid/unpure atomic orbitals of each carbon don't form (or it's never shown as) MOLECULAR orbitals. Why is this?

Here's a lovely video depicting Benzene's structure:

http://www.youtube.com/watch?v=-EFEfVDUp6M

With colorblindness being a recessive "X-linked" (allele is present on on the X chromosome) disorder, a female (XX) must have both recessive colorblind alleles in her genotype in order to express the colorblind phenotype at the organismal level while a male (XY) just requires one copy of the recessive colorblind allele in order to be colorblind due to his Y chromosome's (which is shorter than the X chromosome) inability to have an equivalent locus in order to "mask" the recessive X-linked allele. In regards to specifically a female CARRIER of color blind, meaning she would be heterozygous/monohybrid for the genotype (thus X color blind & X) and thus ideally not express colorblindness, what would occur if (by random chance) the X chromosome WITHOUT the recessive color blind allele is INACTIVATED (via its XIST gene producing a RNA transcript that will eventually, in combination with other factors/proteins, attract methyl/CH3- groups, which will cause condensing of the DNA thus producing "heterochromatin") to produce a Barr Body, and thus leaving only the "activated"/non-supercoiled X chromosome with the recessive color blind allele? Would she be colorblind even though she her genotype is heterozygous? I would greatly appreciate positive responses.

I have not taken a quantum mechanics course yet but I'm curious about the topic. Instead of considering an electron as both a particle and a wave, can't it just be considered as a particle that causes waves in by interacting with space around the nucleus (I'm not sure how it would interact with empty space but the subsequent sentence tries to explain my thought)? I'm trying to picture it as a particle such as a pebble (electron) that interacts with a liquid medium (space around nucleus) and creates waves (electron probability regions/atomic or molecular orbitals). Any (presumably NOT "negative" like an electron) thoughts?

Hello,

The question asks for the "highest" vapor pressure (thus requires the "weakest" intermolecular force).

My options are:

a) NH3 (which would possess H-forces, thus IMF is stronger thus lower vapor pressure)

b) SbH3

c) AsH3

d) PH3

**Out of SbH3, AsH3, & PH3 (which all possess dipole-dipole interactions), I chose "b" (SbH3) primarily because "Sb" (even though it's the largest central element in the given choices) is the LEAST electronegative (thus weaker pull of the hydrogen's electrons) thus I would think that the "partial charges" would be lower. However, the answer was "d" (PH3) even though I think that the dipole-dipole interaction of PH3 will be greater with due to the more electronegative P thus greater pull of electrons from hydrogen thus greater partial charge.

**I guess I have to assume that if all the molecules possess dipole-dipole interaction with just an alternating central atom, the smaller/lower weight central atom (instead of the least electronegative, as I would infer) in combination with the polar molecule will generate a "weaker" dipole-dipole interaction thus increasing the vapor pressure because the forces are easier to break.

Any suggestions?

Hello,

I understand that an Ion-dipole intermolecular force magnitude/strength increases with an increase in charge of the ion. However, the increased charge may be of a "cation" (Example: Aluminum 3+) or of an "anion" (Ex: Phosphorus 3-). The charge is relatively high for both types of ions but which would contribute to a stronger ion-dipole force. I would state that the "cation" (Al 3+) would interact more strongly with a polar molecule (in this case H2O) than the anion P 3- because Al 3+ will possess three less electrons, thus the "Z effective nuclear charge" (the attraction of the + protons in the nucleus to the outer electrons) will increase making the cation/Al 3+ smaller than the P 3- (which will have three extra electrons thus greater repulsions from other electrons will occur with a lower quantity of Zeff). Since the Al 3+ will be smaller, it can get closer to the polar molecule/H2O and according to myself, increase the ion dipole strength. Can anyone verify this?

Hello,

At the moment liquid water at 0 degrees Celsius changes to water/ice at 0 degrees Celsius (thus a phase change), I understand there is no change in kinetic energy/heat because that energy is being used to break the intermolecular forces/H-forces between the water molecules. However, I don't understand why the "potential energy" would decrease during this phase change from liquid to ice/solid water because, since ice is less dense than either of it's previous forms, the distance between the H2O water molecules will be greater thus an INCREASE in potential energy (because H2O forms a net like structure in it's solid state). Any suggestions?

In regards to the theoretical (but very likely) "Higg's Field" (produced from the interaction of Higg's Bosons), why do all other particles (besides photons and gluons) interact with this field (thus giving them mass) and not photons or gluons?

**If do discover the Higg's Field, would it be possible to manipulate it and ultimately turn objects of mass to massless?

I would like to propose my view on this situation. Light travels at speed of "c" universally, however, the path may distorted by the common phenomenon known as refraction. Now, if photons and neutrinos were placed inside the (vacuumed) Large Hadron Collider, we should expect light to win the race. However, even though the circuit is "vacuumed," this doesn't necessarily mean that there's "nothing" there. That "something" may be particles of some sort that may serve to hinder the pathway of the photons (the photons may still be going at the speed of light while bumping into such a small amount of particles/different medium, thus distorting the speed as a whole. Example: If you have to flashlights, and you shine, simultaneously, one of them with the air as the medium and the other with glass as the medium, the one traveling through air would reach an indicated destination before the other.) The neutrinos are very unlikely to interact with any sort of particles due to it's lack of charges, and particles in the vacuum would not hinder/refract it's trajectory. Even though, by stating that there could be small amount of particles still within the vacuum that could hinder a photons directly, seems negligible, the measurement itself was very precise (to the nanometer) leading myself to think that accounting for such measures is logical. I am not a physicist of any sort, but nonetheless intrigued. If I have made some (or several) errors in my explanation, please correct me so that I may adjust or ultimately delete this silly question and view. Thanks a lot.

I would like to propose my view on this situation. Light travels at speed of "c" universally, however, the path may distorted by the common phenomenon known as refraction. Now, if photons and neutrinos were placed inside the (vacuumed) Large Hadron Collider, we should expect light to win the race. However, even though the circuit is "vacuumed," this doesn't necessarily mean that there's "nothing" there. That "something" may be particles of some sort that may serve to hinder the pathway of the photons (the photons may still be going at the speed of light while bumping into such a small amount of particles/different medium, thus distorting the speed as a whole. Example: If you have to flashlights, and you shine, simultaneously, one of them with the air as the medium and the other with glass as the medium, the one traveling through air would reach an indicated destination before the other.) The neutrinos are very unlikely to interact with any sort of particles due to it's lack of charges, and particles in the vacuum would not hinder/refract it's trajectory. Even though, by stating that there could be small amount of particles still within the vacuum that could hinder a photons directly, seems negligible, the measurement itself was very precise (to the nanometer) leading myself to think that accounting for such measures is logical. I am not a physicist of any sort, but nonetheless intrigued. If I have made some (or several) errors in my explanation, please correct me so that I may adjust or ultimately delete this silly question and view. Thanks a lot.