12 / 24.3 x 6.022x1023 = 2.97x1023 atoms
To find the number of nitrogen atoms in 5.00 g of magnesium nitride (Mg₃N₂), first calculate the molar mass of Mg₃N₂, which is approximately 100.95 g/mol (24.31 g/mol for Mg and 14.01 g/mol for N). The number of moles in 5.00 g of Mg₃N₂ is about 0.0495 moles. Since each formula unit of Mg₃N₂ contains 2 nitrogen atoms, the total number of nitrogen atoms is calculated as: 0.0495 moles × 2 × 6.022 × 10²³ atoms/mole, which equals approximately 5.95 × 10²² nitrogen atoms.
There are 2.5 g in 2500 mg, as there are 1000 mg in 1 g.
48 g 19 mg is equal to 48.019 g.
how many moles are represented by 1.51 x 10^24 atoms Pb
Since there are 1,000 milligrams in a gram, 273 g equal 273,000 mg.
1 gram = 1000 milligrams so 12 g = 12*1000 = 12000 mg. Simple!
1 mg = (1/1000) g 1 g = 1000 mg So 12 mg = (12/1000) g Obviously 1 g = 1000 mg which is much greater than 12 mg
1g = 1000mg Therefore, 12g = 12000mg
To find the number of nitrogen atoms in 5.00 g of magnesium nitride (Mg₃N₂), first calculate the molar mass of Mg₃N₂, which is approximately 100.95 g/mol (24.31 g/mol for Mg and 14.01 g/mol for N). The number of moles in 5.00 g of Mg₃N₂ is about 0.0495 moles. Since each formula unit of Mg₃N₂ contains 2 nitrogen atoms, the total number of nitrogen atoms is calculated as: 0.0495 moles × 2 × 6.022 × 10²³ atoms/mole, which equals approximately 5.95 × 10²² nitrogen atoms.
There are 0.120 g in 120 mg. 1 g = 1000 mg 1 mg = 0.001 g
1 g = 1,000 mg 2 g = 2,000 mg 3 g = 3,000 mg . . . 15 g = 15,000 mg
There are 2.5 g in 2500 mg, as there are 1000 mg in 1 g.
1 G = 1,000 mg 0.2 G = 200 mg 0.02 G = 20 mg
1 g = 1,000 mg 2 g = 2,000 mg 3 g = 3,000 mg . . . 100 g = 100,000 mg
48 g 19 mg is equal to 48.019 g.
1 g = 1,000 mg 5 g = 5,000 mg
1,000 mg = 1 g 6,500 mg = 6.5 g