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10.81g

This is found by looking at the atomic weight. The atomic weight refers to the mass of a singe atom in units of AMU (Atomic Mass units) and it also refers to the mass of one mole of the element in grams.

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How many grams are in one mole of boron?

1.00*10-7 moles of boron is how many grams is this


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Each mole of boron atoms has a mass of 10.811 grams, as indicated by the gram atomic mass or weight of boron. Therefore, 585 moles has a mass of about 6.32 X 103 grams, to the same number of significant digits as 585.


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There is no direct relationship between atoms of boronand grams of boron. Use Avogadro's number to convert atoms to moles, and the atomic mass to convert moles to grams.Since you are converting from atoms B, this goes in the denominator (on the bottom) of the first factor. You want to end up in units of grams of B, so this goes in the numerator (on the top) of the last factor.atoms B1.00 mole B10.8 gram = g B6.02E+23 atom B1.00 mole BNote that the units atoms boron "cancel out" in the first factor and you are left in units of moles. Moles cancel outin the second factor and the final units are grams boron.


How many moles are in 97.2 grams of boron?

To do this, you need to know the atomic weight of the element you're dealing with, found on any periodic table. The atomic weight is the mass in grams of the element in one mole - this will provide you with a conversion factor. So take the measurement in grams and divide it by the atomic weight to convert to moles. Really what you're doing is multiplying the number by 1 mole, and dividing it by the equivalent of one mole, the atomic weight. That's the thought process behind unit analysis and how you get your "units to cancel".In this case, the answer is about 8.99 moles B.


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How many grams are in one mole of polonium?

By definition, one mole would be the same as the atomic mass. You take the number of moles and multiply it by the atomic mass. So if you have just 1 mole, the number of grams will be the atomic mass. Polonium's atomic mass is 209 grams.