8.90 X 1023 lead atoms (1 mole Pb/6.022 X 1023)(207.2 grams/1 mole Pb) = 306 grams of lead =============
The mass of 4.21 x 10^23 atoms of phosphorus (P) can be calculated by multiplying the number of atoms by the atomic mass of phosphorus. The atomic mass of phosphorus is approximately 31. Therefore, the mass of 4.21 x 10^23 atoms of phosphorus would be around 1.30 x 10^25 grams.
1 mole has 12.01 grams and has 6.022 x 1023 atoms. There are 6 carbon atoms in a glucose molecule so that times six would give you a total of 72.06 grams out of the 180.156 (molar mass for glucose). Carbon makes up about 40 percent of the total glucose mass so the final answer would be it would be around 2.4088 x 1023 atoms of carbon in one gram. Times that by 4 and you'll get 9.6352 x 1023 atoms of carbon in four gram of glucose.
A gram atom of an element is the mass in grams of Avogadro's Number, about 6.022 X 1023, of atoms of the element.
A gram atom of an element is the mass in grams of Avogadro's Number, about 6.022 X 1023, of atoms of the element.
The molar mass of nitrogen (N) is 14.01 g/mol. One mole of N atoms is 6.022 x 10^23 atoms. Thus, the mass of 6.022 x 10^23 N atoms is 14.01 grams.
8.90 X 1023 lead atoms (1 mole Pb/6.022 X 1023)(207.2 grams/1 mole Pb) = 306 grams of lead =============
To convert atoms to grams, you need to take the number of atoms, divide it by Avogadro's Constant, then multiply it by the atomic mass.Atoms ÷ (6.02 × 1023) × Atomic mass = Mass in grams1.505 × 1023 ÷ (6.02 × 1023) × 12.0 = 3.00 grams Carbon
The mass of 4.21 x 10^23 atoms of phosphorus (P) can be calculated by multiplying the number of atoms by the atomic mass of phosphorus. The atomic mass of phosphorus is approximately 31. Therefore, the mass of 4.21 x 10^23 atoms of phosphorus would be around 1.30 x 10^25 grams.
1 mole has 12.01 grams and has 6.022 x 1023 atoms. There are 6 carbon atoms in a glucose molecule so that times six would give you a total of 72.06 grams out of the 180.156 (molar mass for glucose). Carbon makes up about 40 percent of the total glucose mass so the final answer would be it would be around 2.4088 x 1023 atoms of carbon in one gram. Times that by 4 and you'll get 9.6352 x 1023 atoms of carbon in four gram of glucose.
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
6.022 x 1023= 6.022e+23
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
To find the number of atoms in 5 grams of AuCl3, we first need to determine the molar mass of AuCl3. The molar mass of AuCl3 is 303.325 g/mol. Using Avogadro's number (6.022 x 10^23 atoms/mol), we can calculate that there are approximately 1.65 x 10^22 atoms in 5 grams of AuCl3.
A gram atom of an element is the mass in grams of Avogadro's Number, about 6.022 X 1023, of atoms of the element.
A gram atom of an element is the mass in grams of Avogadro's Number, about 6.022 X 1023, of atoms of the element.