Why is pressure necessary?  I have an understanding of organic chemistry; would it be accurate to guess that pressure is necessary to allow strained tetrahedral carbon-carbon bond angles (ie less than 109.5 degrees)?
135387
pressure adds energy increases the activation energy of the bonding reactions such that a different geometry is achieved
585810
Water, when heated from a solid to liquid state, but ultimately in a gaseous state it again expands.  It still obeys all the laws of chemistry, but the bonds just arrange themselves in a more space efficient manner in liquid than in solid form.\n\nAnd moses with a P, no metals exist with the properties you say.  It is physically impossible, sorry man, but I'm a biochem major and it's my job to know what stuff does under different stresses, and nothing does what you claim.  However, if you do have a material that does this, get in touch with me, we'll talk to the patent office, and we'll license it out to companies for billions of dollars.
557049
There isn't enough energy at near kelvin temperatures for molecules to do all their normal movements and fluctuations.  \n\nThe contrast to this state occurs at any temperature above 'near absolute zero', take for example the nominal temperatures we are comfortable with.  At this temperature there is more than enough energy available for a molecule to bend and vibrate in literally millions of ways....these vibrations occur on the picosecond timescale and for the most part, all accessible ways to vibrate and bend.\n  As the temperature is reduced there are less vibrational modes available and the more high energy ways to vibrate (generally those involving less favourable configurations) start to not occur.  \n\nThe vibrational efficiency and capability of a molecule thus decreases to the point of a few possible vibrational modes as absolute zero is approached.
30904
The answer has to do with symmetry of the molecule and hydrogen bonding.  That should get you going in the right direction
439027
I think you need to spend a few minutes trying to understand the phase diagrams in the chapter.  Each line on the diagram represents a transition from one state to another.  The axes of the diagrams are temperature and pressure.  I'm sure you'll need to learn to interpret these diagrams for an exam, so you might as well start now...
5714
Interesting scenario first of all i would like to comment. Well actually the bond is due to electronic forces. Hence mass has litle effect on the bond strength. Thus i thikn the effect of mass change should be a small one!! unless you change the charge of proton too!!
214500
Carbon. A diamond is carbon put under a lot of pressure. And haven't you ever heard of somebody drawing with "opaper and carbons"?
4123
All are true except that each atom is bonded to four others.  Carbon's valence is four, and so it can only form four bonds--by having double bonds, there is no room to bond to four atoms.  Diamond, on the other hand, has each carbon atom bonded to four other atoms, but no double bonds.\n\nGraphite forms in sheets, with no covalent linking between pifferent sheets.  That is why it is so soft.\n\nthe picture on the second link does not show the double bonds, but if you pick a single direction and regard all parallel bonds in that direction to be double bonds, the valence states will make sense.
29056
Its' probably a rounding error. It would depend on which masses for the elements you use, how precise you want to be.
445444
I think the pressure is actually 2.3E5 Pa.  That would make more sense.
245199
I just passed and AP Chem and I failed to do this problem.\nI think that it couldn't be done. I think the true answer is no reaction.
429913
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
165614
From my understanding, it should not be, unless subjected to stress, such as bending, which forces molecular orientation.
354959
The gentleman who answered above me is incorrect, and you are right. Water does have two lone pairs.\n\nNow to try and answer your question.\n\nThe difference between lone pairs and bonds is obviously that lone pairs are not involved in a bond (think of it as a bridge) with another molecule. Also, it is usually the case that a bond (which contains two electrons) is composed of one electron from one molecule and another electron from the other molecule, though that is not always the case.\n\nWhether or not the electron pair is in a bond or serving as a lone pair, it should not affect the electron configuration. Taking your water example, bonds and lone pairs reject each other equally, so the molecule adopts what is called a tetrahedral formation, since it has four components coming off the Oxygen (2 bonds and 2 lone pairs)\n\nIf you don't know what a tetrahedral conformation is, think of a four sided pyramid with the oxygen in the middle and each electron component (bond or lone pair) pointing toward a corner. It would not matter which component is pointing to which corner, because they would all be about 109 degrees apart. \n\nNow if one of those lone pairs on the oxygen in your water example formed a double bond with one of the hydrogens (which breaks many rules of chemistry, but it's just an example) now you would have 3 components coming off the oxygen (1 single bond, 1 double bond, 1 electron pair) and they would adopt a trigonal planar (2-D triangle) conformation around the oxygen and would each be 120 degrees apart.\n\nSo in your CH2Cl2, you have a central Carbon with four bonds coming off it. One to each Hydrogen and one to eadch Chlorine. Since that makes four components, the Hydrogens and Chlorines will again adopt a tetrahedral conformation around the Carbon. Each bond will repel the others equally and the angular difference will be around 109.5 degrees. It does not matter where the chlorines or hydrogens are since they will be an equal distance apart in any configuration. \n\nIt's hard to think of this concept in three dimensions while you have to draw it in two dimensions, but it gets pretty easy with practice.
34125
Possibly because it's unconventional, in terms of thinking. You have to think about so many things to properly predict the products of a reaction, like kinetics and the regio or stereospecificity of the product.\n\nI, personally, thought Physical Chemistry (a junior year class) was harder, but it just meant I had to work harder.
389412
In the Lewis structure, dots left alone want to pair up with those of another atom.  If there are 5-6-7 dots all around, the atom will accept electrons to fill it last layer.  If there are 1-2-3 dots, the atom will want to give those electrons away.\n\nSo you look at your Lewis structures, and try to pair up alone dots from one giving atom to a receiving atom. \n\nFor your peroxyde, you would get H.   and :.O.: twice so one link between and H and a O, another link between the other O and the other H, and then the two O's link together.\n\nFor you chlorine, there are 7 dots all around each atom, so they will share their un-matched electron in a covalent bond.  \n\nFor the CH4O, their are 4 dots all around the carbon, so you would think that one H will bond to each (that's 4 H then) but to add the oxygen, you have to insert it between a H and the C (that the hydroxyde radical).  It is written CH3H usually to show the -OH polyatomic ion.\n\nFor the C2H4, the carbon has 4 dots all around it, and they will bond to each other.  Each will link to two hydrogen (to make up the four in the molecule), leaving two dots on each C, so this means that there will be one double bond between the two carbon atoms.\n\nFinally, the CH2O os trickyer.  That carbon, with its four dots, will have one H bond to each one dot, so there are two free electrons left.  That oxygen has six dots, two pairs and two free.  This means that each free electron of the O will link up to one of the free electrons on the C, which means that there is one double bond between the carbon and the oxygen.  You'll have H-C-H for the first part and C=O for the second part.\n\nIt's kinda hard to type these :-)  Hope this helps anyway
444067
Hmmm.. The one affecting the boiling point is the intermolecular force..\nAs you can see, most of the compound here is polar. \n\nTherefore all you have to do is if the compound is consisted only of 2 elements (Group 17), you'll have to calculate the electronegativity difference. The more difference between them, the boiling will increase as well.\n\nIf there is more than 2 atoms in a molecule, you'll have to see the dipole moment magnitude (a.k.a polarity). The higher the magnitude, the higher the boiling point will be.\n\n\npeace\nvixklen