Calculate the dipole moment for the hypothetical H+ -- H- "molecule" in Debye units.\nAssume a bond distance of 74.0 picometre.
207606
look at this website to understand....\n\nhttp://www.chemistry.mcmaster.ca/esam/Chapter_7/section_3.html
76946
1.74g should = 17.4g approx.\n\nso,\n\n x  = H (Hydrogen) = 1.008g
476577
Generally the polarity of a molecule will trump its surface area and vandewaal's forces.  The molecular mass of the two molecules is really quite close.  So my bet is on C-O-2
373842
E = hc/wavelength\n\nE = 6.62 10^(-34) * 3 10^8/ 5.95 10^-7 = 3.34 10^(-19)J\n\ni am qite sure
524466
The most polar bond will be the one between atoms that have the biggest difference in electronegativity\n\nF is the most electronegative element there is.\nC has similar electonegativity to S.\nCl and Br are quite close to each other\nSo the biggest difference in electronegativity is between the C and F, thus the most polar bond is C-F in CF4\n\nThe molecular dipole moment depends on the sum of the dipole moments of all the bonds in each molecule.\nCS2 is linear and the dipole moments are cancelling out each other so the total dipole moment is 0\nBr-Cl has only one polar bond (Cl is more electronegative) so it has a dipole moment\nSCl2 has 2 polar bonds and 2 lone pairs in the central atom (S). So the arrangement of the electron groups is tetrahedral and since there are only 2 bonds, there is a net dipole moment\nCF4 the molecule is tetrahedral but has 4 identical polar bonds so the dipole moments cancel out each other and the molecule has no dipole moment\nC2H2 is linear with the dipole moments of the polar C-H bonds canceling out each other (the molecule is H-Ctriple bondC-H) so the dipole moment is 0
360282
Part A\n\nLondon dispersion forces are also called Van der Walls.  All molecules can have London dispersion forces.  If the molecule can also have dipole- dipole forces or hydrogen bonding in it.  These mask the London disperion forces.  So London dispersion forces are measured in non- polar interacting molecules.\n\nPart B\n\nDiple- dipole forces occur between molecules that have partial charges on their atoms.  Partial positive charges are attracted to partial negative forces.  Like forces repel.\n\nPart C\n\nHydrogen bonding occurs between a hydrogen bond acceptor and a hydroden bond donor.  A hydrogen bond acceptor is an electro negative atom(an atom that wants to have more electrons) with a lone pair of electrons.  A hydrogen bond donor is an electro negative atom with a hydrogen bound to it.
526172
D - it's a symmetrical molecule, so even though the individual bonds are polar bonds, the molecule itself has no net dipole
190208
OK... \n\nin CH4 there are 4 C-H bonds that take 1646 kJ... divide by 4 to get energy per C-H bond (call it x)\n\nin C3H8, there are 2 C-C bonds (draw structure) and 8 C-H bonds. so from 4006, subtract 8x (for 8 C-H bonds) so you get answer for 2 C-C bonds. Then divide by 2 to get bond energy of one C-C bond.
157524
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
452210
ICl2 + should be bent like water since I will have two lone pairs and two Cl's on it.\n\nI3- will be linear\n\nSbCl6- will be octahedral
364720
I love these kinds of problems...lol...\n\nFirst you need to find the empirical formula of the compound.  To do this, pretend that you have 100 grams of the compound.  From your elemental analysis, that means that in that 100 grams, you have 65.45 grams of C, 5.49 grams of H and 29.05 grams of O.  Convert each of those amounts into moles by dividing by each elements atomic weight.  That gives you 5.45 moles of C, 5.49 moles of H and 1.80 moles of O.  Now divide all of those by the smallest number.  (What you're trying to find is the ratio of atoms of each element in the formula.).  WHen you do that you find the empirical formula of the compound is C3H3O. From that empirical formula, the molar mass is 55.\n\nNow, go to the gas law part of the problem.  You need to somehow get to the actual molecular weight of the compound.  Using PV=nRT, you can calculate the number of moles of the compound you have.  (Remember to use the right units for each of the numbers you're given.)  I got something like 0.05 moles.\n\nNow, to get the actual molar mass, divide the mass you have in the flask by the number of moles you just calculated.  This gives a molar mass of 110 g/mol.\n\nSince your empirical formula has a mass of 55, the compound has a formula of C6H6O2.\n\nHope this makes sense...
48284
Are you sure you did not make a typo?\nA Hydrocarbon with the formula C7H17 would be very hard to find.  Usually, the number of Hydrogens in a ["normal"] Hydrocarbon can be determined by doubling the number of carbons and adding two.\nIn the case we have something that looks like Heptane (7 carbons), so I would expect there to be 16 Hydrogens (7*2 + 2).  If you align all the carbons in a row and single bond them to eachother to allow for the maxmium number of C-H bonds, where would you place the 17th Hydrogen atom?\n\nIf you did mean C7H17, then the balanced reaction would be,\n4 C7H17 + 45 O2  =  28 CO2 + 34 H2O\n\nHowever, if you did indeed mean Heptane (C7H16), the balanced reaction would be,\nC7H16 + 11 O2  =  7 CO2 + 8 H2O
240991
Just to warn you, if you read the answers you will never benefeit from this problem as you normally would. And learning this material will be more difficult as time continues. Years from now you will wish you tried harder in this class but I am done... heres a good set of ansewers for you to learn from.\n\n\n\n4. for an ideal gas PV=nRT\n    \nmm = molar mass = m/n \nwhere m = mass and n = moles\n\n   therefore n = m/mm\n\n and PV=(m/mm)RT\n     and mm=mRT/PV\n          \nwhere R is the ideal gas constant, T is temp, P is pressure, and V is volume. Plug your stuff in and make sure to cancel out units.\n\n5. Using Dalton's Law of Partial Pressures\nwe know that for P=Pressure\n P(He) + P(O2) = Total Pressure\nP=nRT/V with only n being different between gases for P. Therefore,\n\n     (RT/V)[ n(He) + n(O2)] = Total Pressure.\n\nusing the relationship n=m/mm\n\n    (RT/V) [m(He)/mm(He)  + m(O2)/mm(O2)] = Total Pressure\n\nand you plug everything in and solve for T/V. And using this you can find the pressure of He alone. Let this ration T/V equal A for now.\n\n       Using PV=nRT,   P=mRT/mmV, or P=mRA/mm\n\n6. 6 d electrons. The correct answer also depends on the symmetry and ligands of the complex, but they could just be asking you to count electrons. In an octahedral low spin complex it would have 0 unpaired electrons. In an octahedral high spin complex it would have 1 unpaired electron. In a tetrahedral complex it would have 4 unpaired electrons. And there are about 20 more symmetries but these are the ones people talk about in beginning courses.\n\nps. you can do your own math.
512747
a) Tetrahedral shape. bond angle : 109.5\nNo lone pair on central carbon\n\nb) Linear shape. Bond angle: 180\nNo lone pair on central Be\n\nc) Trigonal pyramdial shape. Bond angle less than 109.5\na lone pair on Cl. 2 oxgen double-bonded and 1 oxygen single-bonded to Cl. 0 formal charge on Cl and the O=Cl. Formal charge on Cl-O is -1.
14010
Find the values of heat of formation (delta H)of each molecule and follow this equation\nyour answer = Ca(OH)2 + H2 - Ca - 2(H2)
403781
I won't do your homework for you, but you should look here:\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1\n\nGood luck. :)
224475
This is best explained with an example.\n\nTake an oxygen atom. Six electrons in the outer shell, so it would to have two more. In a water molecule the two extra electrons are supplied by two hydrogen atoms, which share their electrons with the oxygen. \n\nAt first sight one would expect a shape like this for the molecule:  H-O-H, with the positively charged protons as far away from each other as possible. In the actual water molecule the angle between the two O-H bonds is about 105 degrees. \n\nWhat happens is that the four lone electrons like to be (on average) away from the bonding electrons. To make this energetically possible the water molecule is bend.