A lot!!
850 moles iron (6.022 X 10^23/1 mole Fe)
= 5.12 X 10^26 atoms of iron
1 Fe= 55.85 1 S= 32.07 4 O= 64.00 Molar mass= 151.5304
Fe2O3 + 2Al --> 2Fe + Al2O3Before:50.0g + 50.0g > 0.0g + (not important)159.69(g/mol) + 26.98(g/mol)In mol (before reaction):+0.3131 mol + 1.853 mol (excess)Reaction (used reactant > formed Fe):-0.3131 mol - 0.6262 mol > + 0.6262 mol FeRemaining (= before - used):0.0 mol Fe2O3 + 1.227 mol Al > 0.6262 mol Fe, this should be multiplied by the molar mass of Fe to get mass in grams: 0.6262 (molFe) * 55.85 (g/molFe) = 34.97 = 35.0 g Fe
To calculate the number of atoms in 13.2 mol of copper, you can use Avogadro's number, which is approximately 6.022 x 10^23 atoms per mole. Multiply 13.2 mol by Avogadro's number to get the number of atoms: 13.2 mol * (6.022 x 10^23 atoms/mol) = 7.93 x 10^24 atoms. Therefore, there are approximately 7.93 x 10^24 atoms in 13.2 mol of copper.
1 mol = 6,022 140 857(79).10e23 atoms.0,25 mol is 1,50553521425.10e23 atoms.
4,80 moles of Fe contain 28,9062761136.10e23 atoms.
Molar mass of iron = 55.845g/mol (atomic weight in grams/mol) 1mol Fe atoms = 6.022 x 1023 atoms Fe 10g Fe x 1mol/55.845g x 6.022 x 1023 atoms/mol = 1.078 x 1023 atoms
Since a mole of a metal is generally considered to be Avogadro's Number of atoms of the metal, the answer is 3.5 times Avogadro's Number or 2.1 X 1024 atoms, to the justified number of significant digits.
Amount of Fe = 55.845/55.845 = 1mol There is 1 mol of Fe in a 55.845g sample. 1 mol of Fe contains 6.02 x 1023 atoms (avogadro constant). Therefore there are 6.02 x 1023 atoms in 55.845g of iron.
Molar mass of Fe(NO3)2 is 55.85 + 2(14.00 + 3(16.00)) = 179.85 g/mol Therefore, number of moles of Fe(NO3)2 present is 53.55/179.85 = 0.2977 mol For each molecule of Fe(NO3)2, there are two atoms of nitrogen associated with it. Therefore, there are 0.2977*2 = 0.5954 mol of nitrogen atoms
3.8 g Fe * 1 mol Fe/55.85 g Fe (molar mass) = .0680 mol Fe .0680 mol Fe * 2 mol HBr/1 mol Fe (found in formula Fe+2HBr=>FeBr2+H2)=.136 mol HBr .136 mol HBr*80.912 g HBr/1 mol HBr=11.004 g HBr (or 11 using 2 sig figs) And the mass of H2 that is produced is 0.14 g
3.0 g Fe x 1 mole Fe/55.8 g = 0.054 moles Fe2.0 g S x 1 mole S/32 g = 0.063 moles STherefore there are more molecules of S than of Fe since one mole of either = 6.02x10^23 atoms.
The number of particles (atoms or molecules) in a substance is known as Avogadro's constant and is 6.02 x 1023 particles per mole. The mole is the amount of a substance that contains the same number of particles as the number of atoms in 12 grams of carbon-12. If we consider uranium 238 as the largest isotopic component of natural uranium, and as it consists of atoms not molecules, then 1 kilogram contains 1000/238 moles (1000 grams the mass, 238 the relative atomic mass), ie 4.20 moles. so the number of atoms will be 4.20 x 6.02 x 1023, or 2.529 x 1024 atoms.
the constant Mole (mol): 6.02 x 10^23 are how many atoms you have per mol so the answer can be 7 mol atoms or 6.02 x 10^23 atoms per mol x 7 actual answer is 4.214 X10^24 atoms in 7 mol
1 Fe= 55.85 1 S= 32.07 4 O= 64.00 Molar mass= 151.5304
A mol = 6,022x10^23 atoms 1,5 mol = 9,033x10^23 atoms Always.
Convert 1 gram Fe to moles by dividing by molecular weight, in this case: 55.8 g/mol.So 1 gram = 1/55.8 or 0.0179 moles of Fe. 1 mole of any substance contains Avogadro's number of atoms or 6.022e23. So multiply 0.0179 by 6.022e23 to get:1.079*1022 atoms of Fe (iron)
6,78.1022 molecules of glucose 2,14388229924.1024 iron atoms 1,25.1025 zinc atoms