1 mole S atoms = 32.065 g S (atomic weight in grams)
1 mole S atoms = 6.022 x 1023 atoms S (Avogadro's number)
Convert atoms to moles.
2.01 x 1022 atoms S x (1mol S/6.022 x 1023 atoms S) = 0.0334mol S
Convert moles to mass in grams.
0.334mol S x (32.065g S/1mol S) = 10.7g S
(2.68 x 10^22 atoms) x (238 g U / 6.022 x 10^23 atoms) = 10.59 g of U.
The number least significant figures in the multiplication and division problem is 3 significant figures (2.68). So, following significant figure rules for multiplication, your answer should have 3 significant figures.
The final answer is: 10.6 g U
1 mole S atoms = 32.065 g S (atomic weight in grams)
1 mole S atoms = 6.022 x 1023 atoms S (Avogadro's number)
Convert atoms to moles.
2.01 x 1022 atoms S x (1mol S/6.022 x 1023 atoms S) = 0.0334mol S
Convert moles to mass in grams.
0.334mol S x (32.065g S/1mol S) = 10.7g S
what is the mass in grams of 3.75*10^21 atoms of Li
2.60 X 10^22 atoms of Uranium (1 mole U atoms/6.022 X 10^23)(238.0 grams/1 mole U)
= 10.3 grams of Uranium
3.93 X 10^22 atoms of uranium (1 mole U/6.022 X 10^23)(238.0 grams/1 mole U)
= 15.5 grams of uranium
For natural uranium: 238,02891 grams
The gram atomic mass of Uranium is 238.
238.03 g/mol
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
I think you meant 3.09x10^24 atoms of sulfur. 3.09x10^24 atoms * (1 mol / 6.02x10^32 atoms) * (32 g / 1mol) = 165 grams
A sulfur molecule has the formula S8. Multiply the number of sulfur atoms (8) times the atomic weight of sulfur in grams (32.065g). The molar mass of S8 = 256.52g S8.
100 Hydrogen atoms have an atomic mass of 100.794, 4 Sulfur atoms have an atomic mass of 128.26, and 1 Lanthanum atom has an atomic mass of 138.90547.
93,341,000,000,000,000,000,000,000 or 933.41 x 1023 atoms of arsenic. You just take Avogadro's number (6.022 x 1023) and multiply 6.022 by the number of moles (155) and then multiply the product by 1023.
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
Atomic mass of Ag: 107.9 grams5.00 grams × (6.02 × 1023 atoms) / (107.9 grams) = 2.79 × 1022 atoms Ag
The mass is 2.86 grams but the weight will be 0.028 Newtons.
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
3.40 X 1022 atoms helium (1 mole He/6.022 X 1023)(4.003 grams/1 mole He) = 0.226 grams ===========
I think you meant 3.09x10^24 atoms of sulfur. 3.09x10^24 atoms * (1 mol / 6.02x10^32 atoms) * (32 g / 1mol) = 165 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
A sulfur molecule has the formula S8. Multiply the number of sulfur atoms (8) times the atomic weight of sulfur in grams (32.065g). The molar mass of S8 = 256.52g S8.
Sulfur has relative atomic mass of 32 and oxygen have that of 16. The molar mass of sulfur dioxide is 64 grams per mole. Therefore there is approximately 0.58 moles (37.14/64) of sulfur dioxide in given weight.
4.771 X 1022 atoms neodymium (1 mole Nd/6.022 X 1023)(144.2 grams/1 mole Nd) = 11.42 grams of neodymium --------------------------------------
3.14 g Cu = # atomsTake the known mass of copper multiply it by Avogadro number and divided by the atomic weight.Atomic weight of copper:63.5 g3.14 g Cu (6.02 × 1023 atoms) / (63.5 grams) = 2.98 × 1022 atoms of Copper
The formula N2O5 shows that there are 2/5 as many nitrogen atoms as oxygen atoms in the compound. Therefore, the number of nitrogen atoms required is (2/5)(7.05 X 1022) or 2.82 X 1022 atoms. The gram atomic mass of nitrogen is 14.0067 and, by definition, consists of Avogadro's Number of atoms. Therefore, the mass of nitrogen required to react with the specified amount of oxygen to produce the specified compound is 14.0067 [(2.82 X 1022)/(6.022 X 1023] or 0.656 grams of nitrogen, to the justified number of significant digits.