The molar mass of NaNO3 is 84.9947 g/mol. Multiplying by 0.254 mol gives 21.5886538 g, or correctly sigFig'd, 21.6 g
The molecular mass of sodium nitrate is 84,9947.
Yes, the molar mass of anhydrous sodium sulfate is 142,035 grams.
Using atomic weights for Na = 23 and N=14 and O = 16, one arrives at a mass for 1 mole of NaNO3 of23 + 14 + (3x16) = 85 grams/mole
23 grams
How do you find the mass ratio. And how many grams fluorine is made at decomposition.
The molecular mass of sodium nitrate is 84,9947.
1 mole of Na2SO4 is 142.05 or 142.1 when using 4 significant figures
Sodium nitrate is NaNO3. The percent nitrate is given by :PCT Nitrate = [ ( 14 + 48 ) / ( 23 + 14 + 48 ) ][ 100 ] = [ 62 /85 ][ 100 ] = 72.94 mass percent
Yes, the molar mass of anhydrous sodium sulfate is 142,035 grams.
The mass of silver nitrate is 30,6 g.
Using atomic weights for Na = 23 and N=14 and O = 16, one arrives at a mass for 1 mole of NaNO3 of23 + 14 + (3x16) = 85 grams/mole
The gram atomic mass of sodium is 22.9898, the formula of the least hydrated form of sodium phosphate is Na3PO4.10 H2O, and the gram formula unit mass of this sodium phosphate is 344.09. Therefore, the mass fraction of sodium in this sodium phosphate is 3(22.9898)/344.09 or about 0.20044, and the grams of sodium in 7.2 grams of this sodium phosphate is 1.44 grams of sodium, to the justified number of significant digits.
Ammonium Nitrate = NH4NO3 or N2H4O3 the total mass of a ammonium nitrate molecule is as follows: 2*14u + 4*1u + 3*16u = 28u + 4u + 48u = 80u now, the total mass of Nitrogen in one ammonium nitrate is 2*14u = 28u. Then, we divide 28 (the mass of Nitrogen) by 80 (total mass)= 28/80 = 0.35, which is the ratio for Nitrogen mass divided by total mass. then, we get the 48.5 grams (total mass) and multiply it by this ratio (0.35): 48.5 * 0.35 = 16.975 grams of Nitrogen in 48.5 grams of Ammonium Nitrate.
1170g
23 grams
The formula mass for boron nitrate is 197 grams per mole.
How do you find the mass ratio. And how many grams fluorine is made at decomposition.