Well, firstly you need to know if it is pure gold. Assuming that it is, convert the grams to moles
19.7g Au x 1molAu/197.0gAu = 0.1moles Au
And then, to get the number of atoms in 0.1 moles of gold, multiply it by Avogadro's number, 6.22x10^23 atoms per mole
0.1mol Au x 6.22x10^23atoms/mole = 6.22x10^22
Atomic mass of Ag: 107.9 grams5.00 grams × (6.02 × 1023 atoms) / (107.9 grams) = 2.79 × 1022 atoms Ag
What is the mass, in grams, of 2 × 1012 atoms of potassium?
This depends on the mass of the gold sample.
equation: g= 175 atoms x 1 mol/6.02x10^23 x 196.97g/1 mol ---> 175 atoms of gold weigh 5.73x10^-20 grams.
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023. AuCl3= 303.5 grams5.00 grams AuCl3 / (303.5 grams) × (6.02 × 1023 atoms) = 9.92 × 1021 atoms
Atomic mass of Ag: 107.9 grams5.00 grams × (6.02 × 1023 atoms) / (107.9 grams) = 2.79 × 1022 atoms Ag
Gold has a molar mass of 196.96655 grams per mole. 100 grams then is equal to .508 moles which makes 3.059 E23 atoms of gold.
5.0 grams gold (1 mole Au/197.0 grams)(6.022 X 1023/1 mole Au) = 1.5 X 1022 atoms of gold ===================
What is the mass, in grams, of 2 × 1012 atoms of potassium?
This depends on the mass of the gold sample.
This depends on the mass of the gold sample.
equation: g= 175 atoms x 1 mol/6.02x10^23 x 196.97g/1 mol ---> 175 atoms of gold weigh 5.73x10^-20 grams.
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023.5.0 grams Fe / (55.9 grams) × (6.02 × 1023 atoms) = 5.38 × 1022 atoms
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023.3.86 grams S / (32.1 grams) × (6.02 × 1023 atoms) = 7.24 × 1022 atoms
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023.1000 grams C / (12.0 grams) × (6.02 × 1023 atoms) = 5.02 × 1025 atoms
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
For this problem, the atomic mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023. AuCl3= 303.5 grams5.00 grams AuCl3 / (303.5 grams) × (6.02 × 1023 atoms) = 9.92 × 1021 atoms