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Amount of Ca = 139/40.1 = 3.47mol

1 mol of Ca contains 6.02 x 1023 atoms (avogadro constant).

3.47mol of Ca thus contains 3.47 x 6.02 x 1023 = 2.09 x 1024 atoms

Q: How many atoms are in 139 g of calcium?

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oxygen

Get off Wiki Answers Mrs. Z's chemistry class

Quite a few! 147 grams calcium (1 mole Ca/40.08 grams)(6.022 X 1023/1 mole Ca) = 2.21 X 1024 atoms of calcium ----------------------------------------

The gram atomic mass of calcium is 40.08. Therefore, 127 g of calcium constitutes 127/40.08 or 3.69 gram atomic masses. By definition of Avogadro's Number, each gram atomic mass contains Avogadro's Number of atoms. Therefore, the answer is 3.69 X Avogadro's Number or 1.91 X 1024 atoms, to the justified number of significant digits.

The formula is CaF2 so, its three atoms, one calcium and two fluorine atoms. add the molar mass of calcium(40 g/mol) and fluorine(19 g/mol X 2 because there are two atoms of fluorine) and you get 40+(19x2)=78 78 grams per mol

Related questions

149 g of calcium contain 22,39.10e23 atoms.

169 g of calcium contain 25,394.10e23 atoms.

61,5 g calcium contain 9,241.10e23 atoms.

When calculating calcium you must first see how many moles are there. There are 2.6Ã?10^24 atoms in 173 grams of calcium.

175 g Ca (1 mol / 40.078 g) (6.022x10^23 atoms / 1 mol) = 2.63x10^24 Ca atoms There are 2.63x10^24 c.alcium atoms in 175 grams of calcium.

There are 6.023 x 10^23 atoms in each mole of calcium. I mole of calcium is 40 g, so 1 g contains 6.023 x 10^23/40 atoms = 1.506 x 10^24

oxygen

Get off Wiki Answers Mrs. Z's chemistry class

The answer is 0,068 mol (for O not for O2).

Quite a few! 147 grams calcium (1 mole Ca/40.08 grams)(6.022 X 1023/1 mole Ca) = 2.21 X 1024 atoms of calcium ----------------------------------------

(g) -> moles = x(g) * (1 mol/molar mass)Moles -> atoms = x(mol) * (6.022*10^23/1 mol)

The gram atomic mass of calcium is 40.08. Therefore, 127 g of calcium constitutes 127/40.08 or 3.69 gram atomic masses. By definition of Avogadro's Number, each gram atomic mass contains Avogadro's Number of atoms. Therefore, the answer is 3.69 X Avogadro's Number or 1.91 X 1024 atoms, to the justified number of significant digits.