Here I will assume that you meant 2.62 X 1066.
1 mole of water weights 18.01 grams
1 mole = 6.023X1023 particles
2.62 X 1066 particles = x moles
6.023X1023 particles = 1 moles
x=4.35 X 1042 moles
then
1 mole = 18.01 grams
4.35 X 1042 moles = y grams
y= 7.83 X 1043 grams
I have a feeling you did a mistake with the number of atoms because this is a very large amount of water but you get the idea on how to solve the problem.
.01456 g (6.02 x 1023 atoms) / (39.1 g) = 2.24 x 1020 atoms
6.81 grams copper (1 mole Cu/63.55 grams)(6.022 X 1023/1 mole Cu) = 6.45 X 1022 atoms of copper ----------------------------------------
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
25,1 grams sulfur contain 4,7147765.10e23 atoms.
The answer is 9,17233.1023 atoms.
1.659 [grams] / 47.998 [grams / mol] * 6.02214179(30)×1023 [molecules / mole] * 3 [atoms / molecule]. A bunch.
3.65 grams of water is equal to .203 moles of H2O. This means there is also .203 moles of H2 present, or .408 grams.
12.044*10^23 atoms 1.5055*10^23 S8 molecules
2.408 x 10^24 atoms.
.01456 g (6.02 x 1023 atoms) / (39.1 g) = 2.24 x 1020 atoms
Five iron atoms have a mass of 1.3155 x 10-22 grams.
6.81 grams copper (1 mole Cu/63.55 grams)(6.022 X 1023/1 mole Cu) = 6.45 X 1022 atoms of copper ----------------------------------------
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
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
50 grams CaCO3 (1 mole CaCO3/100.09 grams)(6.022 X 1023/1 mole CaCO3) = 3.0 X 1023 atoms of calcium carbonate =============================
904,000,000,000,000,000,000,000 (9.04 X 10^23) Bromine atoms.
25,1 grams sulfur contain 4,7147765.10e23 atoms.