I assume you mean 6.2 x 1022, but the logic is the same for other amounts anyway.
6.2 x 1022 atoms is 0.01295 moles (using Avogadro's number)
1 mole gold weights 197 grams (many periodic tables list this in the top right corner)
Therefore your gold weights 197 x 0.01295 = 2.6 g
To find the number of atoms in 100 grams of gold, you would first calculate the number of moles using the molar mass of gold (197 grams/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, approximately 3.01 x 10^23 atoms of gold are present in 100 grams.
5.0 grams gold (1 mole Au/197.0 grams)(6.022 X 1023/1 mole Au) = 1.5 X 1022 atoms of gold ===================
There are about 6.02 x 10^23 atoms in 16.3 grams of sulfur, which is Avogadro's number representing one mole of atoms or molecules. This is known as a mole, where one mole of any element contains Avogadro's number of atoms.
To convert uranium atoms to grams, you need to know the molar mass of uranium. Uranium's molar mass is approximately 238.03 g/mol. Therefore, 7.5 x 10^21 uranium atoms would weigh approximately 0.625 grams.
The formula unit for gold is a single atom, and the atomic weight of gold is about 197. Therefore, the number of atoms in 3.50 g of gold is Avogadro's Number X (3.50/197) or about 1.84 X 1022.
There are approximately 1.93 x 10^23 atoms in 300 grams of gold. This is calculated using Avogadro's number (6.022 x 10^23 atoms/mole) and the molar mass of gold (197 grams/mole).
To find the number of atoms in 100 grams of gold, you would first calculate the number of moles using the molar mass of gold (197 grams/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, approximately 3.01 x 10^23 atoms of gold are present in 100 grams.
Atomic mass of Ag: 107.9 grams5.00 grams × (6.02 × 1023 atoms) / (107.9 grams) = 2.79 × 1022 atoms Ag
There are about 2.56 x 10^22 atoms in 10 grams of gold. This calculation is based on the atomic mass of gold (197 g/mol) and Avogadro's number (6.022 x 10^23 atoms/mol).
5.0 grams gold (1 mole Au/197.0 grams)(6.022 X 1023/1 mole Au) = 1.5 X 1022 atoms of gold ===================
Every drop of seawater contains around 0.12 billion gold atoms.
The number of atoms in a bar of gold depends on the mass of the bar. To calculate the number of atoms, you would need to know the mass of the bar and the atomic mass of gold. You can use Avogadro's number to convert from grams to atoms.
To calculate the number of gold atoms in one kilogram, we first need to determine the number of atoms in 1 gram, then multiply by 1000 to account for the kilogram. Given that one atom of gold weighs 3.27x10^-22 grams, the number of atoms in 1 gram would be 1/(3.27x10^-22) = 3.06x10^21 atoms. Therefore, in one kilogram (1000 grams) of gold, there would be 3.06x10^24 atoms.
To find this, you need to first calculate the molar mass of gold (Au), which is 196.97 g/mol. Next, you convert the given mass (5 grams) to moles by dividing by the molar mass. Then, you use Avogadro's number (6.022 x 10^23) to convert moles to atoms. In this case, there are approximately 6.022 x 10^22 atoms in 5 grams of gold.
To find the number of gold atoms in 45.0 grams of gold, first determine the molar mass of gold (Au), which is approximately 197 g/mol. Next, calculate the number of moles in 45.0 grams by dividing the mass by the molar mass: ( \frac{45.0 \text{ g}}{197 \text{ g/mol}} \approx 0.228 \text{ mol} ). Finally, multiply the number of moles by Avogadro's number ((6.022 \times 10^{23} \text{ atoms/mol})): ( 0.228 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 1.37 \times 10^{23} \text{ atoms} ). Thus, there are approximately (1.37 \times 10^{23}) gold atoms in 45.0 grams of gold.
There are approximately 31.1 grams in 1 ounce of gold.
To calculate the number of atoms in 197 kg of gold, you would first convert the mass of gold to moles using the molar mass of gold, which is approximately 197 g/mol. Then, you would use Avogadro's number (6.022 x 10^23 atoms/mol) to find the number of atoms in the moles of gold.