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Nickel is 58.7g/mol, so 26.4/58.7=0.45mol, and that times 6.02x1023 (Avogadro's number) gives you 2.71x1023 atoms of nickel.
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What is the mass in grams of 5.97 x 1027 atoms of nitrogen?
500,000 atoms * 1 mole Ni/6.022*1023 atoms * 58.693 grams/1 mole Ni = 4.873*10-17 grams Therefore, the mass of 500,000 atoms of nickel is about 4.87*10-17 grams
How many moles of Li atoms are in 10.56grams of Li?
How many lithium atoms are in 10.56 g of lithium
How many atoms of sulfur are in 1 g?
1.87810*10^22 Sulfur Atoms.
How many atoms of lithium are in 11.8 g?
1.07x10
How many atoms of lithium are in 12.0 g?
2.32
How many moles of nickel (Ni) atoms are in 125 g Ni?
There are 58.6934 gram in one mole of Ni atoms, so there are125(g Ni) / 58.6934 (g.mol-1 Ni) = 2.13 moles in 125 gram Ni
How many atoms are in .23 g Pb?
The number of atoms of lead is 6,68.10e23.
How many moles of nickel (Ni) are there in 125 g of Ni?
125 g nickel is equivalent to 2,13 moles.
What is the number of atoms per unit cell for nickel?
According to the U.S. mint the nickels currently in circulation weigh 5.000 grams and contain 25% Ni (the rest is copper). 25% of 5.000g = 1.250 grams of Ni in a nickel coin / molar mass of Ni 58.71 g/mol = 0.02129 moles of Ni times the number of particles for one mole 6.02x10^23 = 1.28x10^22 atoms of Ni in one coin. All together: (0.25 x 5.000g) / (58.71 g per mol) x (6.02x10^23 atoms per mol) = 1.28x10^22 atoms Ni per coin.
How many atoms are in 49.1740 g of zirconium?
49.1740 g (6.02 x 1023 atoms) / (91.22 g) = 3.25 x 1023 atoms
How many atoms are in 149 g of calcium?
149 g of calcium contain 22,39.10e23 atoms.
What is the mass in grams of 5.97 x 1027 atoms of nitrogen?
500,000 atoms * 1 mole Ni/6.022*1023 atoms * 58.693 grams/1 mole Ni = 4.873*10-17 grams Therefore, the mass of 500,000 atoms of nickel is about 4.87*10-17 grams
How many atoms are in 38.260 g of chlorine?
6,687.1023 chlorine atoms
How many atoms are there in 63.5 g of carbon monoxide?
27,30.10e23 atoms
How many atoms are in g Pb?
The number of atoms of lead is 6,68.10e23.
How many atoms are there in 30.4 g of selenium?
78.96 g of selenium will have 6.023 x 1023 atoms. So, 30.4 g of selenium will have 2.32 x 1023 atoms.
How many atoms of sulfur (S) are in 5 g of sulfur?
5 g of sulfur contain 0,94.10e23 atoms.
