#300
12 g of carbon will have 6.023 x 1023 atoms of carbon
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
divide by the atomic mass and times it by advogadro's number.
If the density of carbon tetrachloride is 1.59 g/L, the volume in L of 4.21 kg of carbon tetrachloride is about 2,647.8 L
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
27,30.10e23 atoms
0,515 g of carbon is equal to 0,043 moles.
12 g of carbon will have 6.023 x 1023 atoms of carbon
1 mole of carbon (or 12 g) has 6 x 1023 atoms. So, 3 moles of carbon (or 36 g) has 18 x 1023 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
Atomic mass of carbon: 12.0 grams1.90 grams × (6.02 × 1023 atoms) / (12.0 grams) = 9.53 × 1022 atoms C
divide by the atomic mass and times it by advogadro's number.
30 g of ethane will have 6.023 x 1023 molecules of ethane So, 5.5 g will have 1.104 x 1023 molecules of ethane Since there are two carbon atoms, in one molecule of ethane, 5.5 g of ethane will have 2.208 x 1023 atoms of carbon.
If the density of carbon tetrachloride is 1.59 g/L, the volume in L of 4.21 kg of carbon tetrachloride is about 2,647.8 L
This value is 0,05678.10 ex.23 atoms.
how many moles of CF4 are there in 171 g of CF 4
21.7 g CCl4 (1 mole CCl4/153.81 g)(6.022 X 1023/1 mole CCl4) = 8.50 X 1022 atoms of carbon tetrachloride ===========================