(1.4/Atomic Mass of carbon) X Avogadro's number = 1.4/12.0 X 6.02 X 1023 = 7.0 X 1022 atoms, to the justified number of significant digits.
The number of atoms is 28,099.10e23.
Carbohydrates do not contain "g" atoms. Thus we can not answer your question.
1 mole of the element has 6.023 x 1023 atoms 1 mole = 4 g of helium = 7 g of lithium = 9 g of beryllium = 11 g of boron = 12 g of carbon = 14 g of nitrogen = 16 g of oxygen
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.
156,86.1020 atoms of uranium in 6,2 g uranium
1.6 g C x 1 mol C/12 g x 6.02x10^23 atoms/mol = 8.1x10^22 atoms
85.9 (g C) = 85.9 (g C) / 12.00 (g/mol C) = 7.158 (mol C)7.158 (mol C)*[6.022*1023 (atoms/mol C)] = 4.31*1024 C-atoms
169 g C x 1 mole C/12.011 g x 6.02x10^23 atoms/mole = 8.47x10^24 atoms
The number of atoms is 28,099.10e23.
Since 12 g C is 1.0 mole C there are 6.02*10+23 atoms in (each) mole; that's the definition of a mole
Carbohydrates do not contain "g" atoms. Thus we can not answer your question.
One mole of any element contains 6.02 x 1023 atoms. So 1mole of carbon = 6.02x1023 atoms. 0.8mole of carbon = 0.8 x 6.02x1023 0.8 mole of carbon = 4.816 x 1023
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
The number of atoms of lead is 6,68.10e23.
divide by the atomic mass and times it by advogadro's number.
1 mole of NaHCO3 Na = 1 * 22.99 g = 22.99 g H = 1 * 1.01 g = 1.01 g C = 1 * 12.01 g = 12.01 g O = 3 * 16.00 g = 48.00 g Total = 84.01 g There are 84.01 grams in one mole of NaHCO3.
25.0 g C10H8N2O2S2 ( 1 mole C10H8N2O2S2/252.324 g)(10 mole C/1 mole C10H8N2O2S2)(6.022 X 1023/1 mole C) = 5.97 X 1023 atoms of carbon ---------------------------------------