6.02 x 10^23
2007-10-22 09:24:10
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answer #1
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answered by Dastardly 6
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Number of atoms in a mole depends on how many atoms are in each molecule of the compound, unless you are measuring moles of atomic substances. Units per mole is 6.023x10^23, so atoms per mole of a molecular substance is that times the number of atoms per molecule (Oxygen would be 2x6.023x10^23)
2007-10-22 09:27:46
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answer #2
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answered by vyperjeedai 4
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The mole (symbol: mol) is the SI base unit that measures an amount of substance. One mole contains Avogadro's number (approximately 6.023×10^23) entities.
A mole is much like "a baker's dozen" in that both are absolute numbers (having no units) and can describe any type of elementary object. The mole's use, however, is usually limited to measurement of subatomic, atomic, and molecular structures; tradition and its hugeness compared to more common units make it impractical for other uses.
2007-10-22 09:23:42
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answer #3
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answered by DanE 7
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Avagadro's number: 6.02 x 10^23, either atoms or molecules, are in a mole.
2007-10-22 09:25:51
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answer #4
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answered by JB 2
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A mole is 6.02 * 10^23 pieces of anything. In the case of a compound, it's that number of molecules. In the case of an element, it's that number of atoms.
The atomic weight of an element tells you how many grams are in a mole of that element.
2007-10-22 09:24:38
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answer #5
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answered by Brent L 5
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Remove Your Moles, Warts & Skin Tags in 3 Days
Easily, Naturally And Without Surgery
2016-05-17 12:27:16
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answer #6
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answered by Anonymous
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number of molecules in a mole= Avogadroes Number
=6.03x10^23
2007-10-22 09:25:31
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answer #7
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answered by Anonymous
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RE:
how many atoms are in a mole? how many molecules are in a mole?
its mole day tommorow and i neeed to know!
2015-08-02 02:42:59
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answer #8
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answered by Anonymous
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1
2017-02-28 23:39:47
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answer #9
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answered by Robinson 3
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Avogadro number
6.023×10^23
2007-10-22 09:24:25
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answer #10
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answered by Erunion11 2
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Exactly 152,5133,6112 Molecules are in a mole
hummmmm or was it 125,5133,6221 i dont remember
2007-10-22 09:25:44
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answer #11
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answered by shikage9999 2
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