Where would the chiral centres be located on a 3D glucose molecule?
823978
there is only one chiral center.  steriogenic centers are when there are 4 different groups on the same C.
436740
CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle
62858
The difference between fructose and glucose is that the aldehyde (the lone oxygen) is attached to the end carbon in glucose but on the second carbon in fructose.
742243
Dont you mean how a molecule of glucose releases energy in a cell....any which way you look at it...refer to Kreb's Cycle in your Biology textbook!
419119
The chiral organization of molecules (and other matter) is a frequently occurring phenomenon in our universe.  In Earth's early history, conditions were right (a large variety of atoms in the primordial soup, perfect temperature etc) to produce billions and billions of chirally organized combinations of molecules.   All that is needed for life to begin is a charged lipid micelle (common in Earth's early oceans) to absorb just one correctly organized chain and voila - the chain now has its closed environment complete with power source in which to sustain the electro-chemical reactions needed for life
947336
D - it's a symmetrical molecule, so even though the individual bonds are polar bonds, the molecule itself has no net dipole
41506
We're talkin about a polar molecule when the center of gravity of its negative charges is different from the center of gravity of its positive charges. The most famous example of a polar molecule is H20\n .......H (+)\n......./ &#92;\n..(-)O...O (-)\n\n(NB: the dots are only here for the simplicity, don't pay attention to them)\nOn this very shematic drawing, you can see that the center of gravity of the (partial)positive charges is located on the Hydrogen atom whereas the center of gravity of the negative charges is located half way between the 2 oxygen atoms.\nOn the other side, a non polar molecule, such as CO2 (see below)\n\n (-)O=C(+)=O(-)\n\nhas the center of gravity of both its negative and positive charges at the same place, i.e. on the carbon atom.\nFor the economic size, I won't be able to halp ya, sorry, I'm just a scientist, However, the polarity of a molecule is of great interest in a lot of domain (dielectrics, solvants...)
583606
maltose it is glucose+glucose
260431
R-O-CH3 \n\n R is any carbon based group. I wish I had chem draw on this computer, but the structure is actually bent due to the symmetry (this involves group theory - a linear algebra based idea in quantum mechanics) of the orbitals involved in bonding. Its bent for the same reasons that water is bent, now your teachers may say that water is bent due to electronic repulsions but if thats true how come they draw the lone pairs so close to eachother??? Its because its not true, and its part of VSEPR theory which is just a practicle method of teaching structure to people who dont know too much physics, math, and chemistry.
922867
Check out that link.  It has a picture that shows how it looks like: the grey Al atom is in the center, and 3 green Cl atoms are around it in a triangular arrangement.
956604
Try it out in the lewis-dot diagram\n      ..        ..\nH-O-Ca-O-H\n      ..        ..
284075
It's fairly easy don't worry.\n\nNo need for d level splitting diagrams to answer this since we're dealing with N and F:  use the spectrochemical series to make an assumption about the ligand.  The difference in t2g and eg levels (∆o) determines whether a complex is low or high spin.  If both ligands were the same, we would have to look at the oxidation state of the ligand in the complex.\n\nHalides <  Oxygen ligands  <  Nitrogen ligands < CN- ligands\n\n   ALWAYS HIGH SPIN    ---------------            ALWAYS LOW SPIN\n\nIf ∆o > P, then a complex is low spin.  IF ∆o < P the complex is high spin.
852575
achiral.  you need to have four different functional groups on any given carbon center to have a chiral molecule.  First, by default CH3 groups are not chiral (because they always only have two different groups - three H's and whatever else is attached to it).  That leaves two other carbon centers which have the following number of different groups:\n1.  2 - three CH3 groups and one CH(CH3)2\n2.  3 - one (CH3)3C group, one H group, and two CH3 groups\n\nSince there are no carbon centers with four different functional groups, the molecule is not chiral.  Hope this helps
952829
In all molecules, there is a type of chemical energy called "enthalpy", it is the energy stored in bonds.  Think of it this way, all particles want to be free and in motion, in complete disorder, thats the "entropic" nature of the universe.  So, it takes energy to make particles stick together (forming compounds and molecules).  This energy is stored in bonds, and when the bonds are broken, the energy is released.  Our cells have the ability to break these bonds, and utilize this energy.\n\nThink of it this way, glucose is C6H12O6.  Our cells can break down all of the bonds in glucose until we form a few CO2 and water molecules.  Thats how we get our energy.
789565
a) 4 "things" attached to central atom and no lone pairs then: sp3-> tetrahedral e- arrangement, tetrahedral shape, 109.5°\nb) 4 "things" attached to central atom one lone pair then: sp3 -> tetrahedral e- arrangement, trigonal pyramidal shape, < 109.5° (about 107°)\nc) 3 "things" attached to central atom one lone pair then: sp2 -> trigonal planar e- arrangement, bent shape, < 120° (guess about 115°?). This is most challanging, try to figure out how many e- in total you have (18), then arrange them around the central atom using usual valances.
734320
In terms of chemistry, mixtures and compounds are irrelevant to each other.\n\nThink of it this way... the "opposite" of a compound is an element. Glucose is not an element (C6H12O6), so it must be a compound.\n\nTo: thekevinzhang: glucose is absolutely not an ionic compound. Don't talk when you have no clue what you're talking about.
711569
If the ligands lie in the x-y plane, then the d-orbitals on the central atom that have all their electron density *out* of this plane will have the lowest energies.  These are the d-xz and d-yz orbitals. \n\nThe d-z^2 orbital has its density mostly out of the x-y plane, but there is a ring of electron density in the x-y plane, so this orbital will have a somewhat higher energy than the d-xz and d-yz orbitals.\n\nThe d-xy and d-(x^2-y^2) orbitals have their electron density concentrated in the x-y plane so their energies will be the highest.  They have the same energy.\n\nSummary, highest energy at the top:\n\nd-xy, x-(x^2-y^2)\nd-z^2\nd-xz, d-yz