1.5g/12.0107g = n,
n * 6.022*10^23 = 7.520793959e22
In this case n would equal .124888641
12.0107 is the grams in one mole of Carbon (Found in the Periodic Table -- Atomic Mass)
6.022*10^23 is avogadro's number or one mole.
1.6 g C x 1 mol C/12 g x 6.02x10^23 atoms/mol = 8.1x10^22 atoms
There are 7.16 moles of carbon in 85.9 g of carbon (85.9 g / 12 g/mol). Since there are 6.022 x 10^23 atoms in 1 mole of carbon, the number of carbon atoms in 85.9 g is 4.31 x 10^24 atoms.
To find the number of carbon atoms in 12 g of carbon (C), you first need to calculate the number of moles of carbon (C) in 12 g using its molar mass. Then, you can use Avogadro's number (6.022 x 10^23) to determine the number of atoms in that many moles of carbon (C).
The number of atoms of lead is 6,68.10e23.
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 ---------------------------------------
1.6 g C x 1 mol C/12 g x 6.02x10^23 atoms/mol = 8.1x10^22 atoms
There are 7.16 moles of carbon in 85.9 g of carbon (85.9 g / 12 g/mol). Since there are 6.022 x 10^23 atoms in 1 mole of carbon, the number of carbon atoms in 85.9 g is 4.31 x 10^24 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.
To find the number of carbon atoms in 12 g of carbon (C), you first need to calculate the number of moles of carbon (C) in 12 g using its molar mass. Then, you can use Avogadro's number (6.022 x 10^23) to determine the number of atoms in that many moles of carbon (C).
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 ---------------------------------------