September 3 Notes

September 3 Notes - Chemistry 151 September 3, 2008...

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Chemistry 151 September 3, 2008 Avogadro’s Number: 6.0221420•10 23 Avogadro’s Number is equal to the number of atoms in 12 g (exactly) of 12 6 C. We call 6.022•10 23 particles of something one mole (abrv. mol) of that substance. One mole of an elemental substance has a mass equal to that element’s average atomic mass expressed in grams. Mass Mole number of particle conversions 1 amu=1.66054x10 -24 g 1 g = 6.022x10 23 amu 1 mol = 6.022x10 23 particles To calculate the average atomic weight of an element, take the weighted average. o Multiply weight by percent abundance. The Periodic Table Dmitri Mendeleev (1834-1907) Organizing scheme for the elements. o Based on increasing mass. Also noticed elements in regular intervals share similar properties. o Vertical columns = groups We now know that rather than by increasing mass, elements should be ordered on increasing atomic number (number of protons in the nucleus). Organization
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This note was uploaded on 12/01/2009 for the course CHEM 151 taught by Professor Metcalf during the Fall '05 term at UVA.

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September 3 Notes - Chemistry 151 September 3, 2008...

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