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DrB
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Marcia I don't know why it is wrong. But I'm sure about the answer. The atoms you need broke on (100) plane are the four in the plane bonded to the body central atom (remember that each atom in a BCC Crystal has 8 neighbors but you only need broke 4 - if
I don't know why it is wrong. But I'm sure about the answer. The atoms you need broke on (100) plane are the four in the plane bonded to the body central atom (remember that each atom in a BCC Crystal has 8 neighbors but you only need broke 4 - if you
because I don't think this is the right answer. A racemic mixture in a not readily biodegradable chemical with high toxicity???? hmmmmmm 1,1,2-Trichloroethane is produced in closed systems and used as a solvent...???
The enthalpy of atomization of nobium is 745 KJ/mole. The energy per atom is = 745000/Avogadro Number = 1.2371305e-18 J/atom In a BCC structure each atom has 8 neighbors. Then The energy per bond is = 1.2371305e-18 / 8 = 1.5464132e-19 J/bond Now in a (100)
The enthalpy of atomization of nobium is 745 KJ/mole. The energy per atom is = 745000/Avogadro Number = 1.2371305e-18 J/atom In a BCC structure each atom has 8 neighbors. Then The energy per bond is = 1.2371305e-18 / 8 = 1.5464132e-19 J/bond Now in a (100)
The enthalpy of atomization of nobium is 745 KJ/mole. The energy per atom is = 745000/Avogadro Number = 1.2371305e-18 J/atom In a BCC structure each atom has 8 neighbors. Then The energy per bond is = 1.2371305e-18 / 8 = 1.5464132e-19 J/bond Now in a (100)
The enthalpy of atomization of nobium is 745 KJ/mole. The energy per atom is = 745000/Avogadro Number = 1.2371305e-18 J/atom In a BCC structure each atom has 8 neighbors. Then The energy per bond is = 1.2371305e-18 / 8 = 1.5464132e-19 J/bond Now in a (100)
bonjo. Are you hunting ducks in the dark???
The enthalpy of atomization of nobium is 745 KJ/mole. The energy per atom is = 745000/Avogadro Number = 1.2371305e-18 J/atom In a BCC structure each atom has 8 neighbors. Then The energy per bond is = 1.2371305e-18 / 8 = 1.5464132e-19 J/bond Now in a (100)
DrBob222 is SOO close... Yes, the ratio is 1:1, but how does the molecular mass of HgBr compare to the molecular mass given in the problem (this will give you the answer)?
