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Divide 1.8*10^20 (unit: number) particles by 6.02*10^23 (unit: number per mole), Avogadro's number) and you'll get

3.0*10^(-4) moles of those particles

(In your case particles are atoms, but this is also valid for ANY kind of particles)

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June Douglas

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3y ago

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Related Questions

How many atoms are there in silver?

A mole of silver contains approximately 6.022 x 10^23 atoms.


How many silver atoms are there in 8.68 moles of silver?

There are 6.022x1023 atoms in a mole. You multiply 6.022x1023 by 8.68, which equals 52.20796x1023 atoms


How many grams are there in 2.3x10 to the 24th power atoms of silver?

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How many atoms are in 0.050 mol of silver?

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What can be said about 1 mole silver and 1 mole gold?

1 mole of silver contains Avogadro's number of silver atoms, while 1 mole of gold contains Avogadro's number of gold atoms. The molar mass of silver and gold can be used to determine the mass of each element in 1 mole. Both contain the same number of atoms per mole due to Avogadro's number.


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There are 4 O atoms per molecule Ag3PO4 .In 0.02 mole Ag3PO4 are0.02(mole) * 6.022*10+23(molecules/mole) * 4(atolms/molecule) = 4.8*10+22 atoms Oxygen


How many atoms of silver in 10.8g?

To find the number of atoms of silver in 10.8g, you need to first calculate the number of moles of silver using its molar mass. Then, you can use Avogadro's number (6.022 x 10^23 atoms/mol) to convert the moles of silver to number of atoms.


How many silver atoms are there in 3.76g of silver?

To calculate the number of silver atoms in 3.76g of silver, you need to use Avogadro's number and the molar mass of silver. The molar mass of silver is 107.87 g/mol. First, calculate the number of moles in 3.76g of silver. Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to find the number of silver atoms in that many moles.


How many silver atoms are contained in 3.75 moles of silver?

There are 2.26 x 10^24 silver atoms in 3.75 moles of silver. This is calculated by multiplying Avogadro's number (6.022 x 10^23 atoms/mole) by the number of moles.