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What is the mass of 6.0 moles of sodium nitrate (NaNO3)?
Cheyenne Hogue
1 mole of Na2SO4 is 142.05 or 142.1 when using 4 significant figures
Wiki User
∙ 14y ago
Wiki User
∙ 10y agoSodium nitrate molecular weight is 84,9947.
aronmitchell
509.96 grams
Cortney Runolfsdotti...
Nadia Heidenreich
Felipe Marvin
Anais Lang
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Which weighs more 5moles of sodium nitrate or 4 moles of carbon dioxide?
I'm assuming that your instructor is using the terms mass and weight interchangeably. First determine the molar mass of each compound by multiplying the molar mass of each element by its subscript and add them together. The molar mass of an element is its atomic weight on the periodic table in grams. Molar mass of sodium nitrate (NaNO3) (1 atom Na x 22.98970g/mol Na) + (1 atom N x 14.0067g/mol N) + (3 atoms O x 15.9994g/mol O) = 84.9946g/mol NaNO3 Molar mass of carbon dioxide (CO2) (1 atom C x 12.0107g/mol C) + (2 atoms O x 15.9994g/mol O) = 44.0095g/mol CO2 Now multiply the molar mass for each compound by the number given and compare the results to determine which has the greater mass. 5mol NaNO3 x 84.9946g/mol NaNO3 = 424.973g NaNO3 4 mol CO2 x 44.0095g/mol CO2 = 176.038 So 5 moles of sodium nitrate has a mass greater than 4 moles of carbon dioxide.
What is the percentage by mass of nitrogen in sodium nitrate?
The formula for sodium nitrate is NaNO3, showing that each formula unit contains one sodium atom, one nitrogen atom, and three oxygen atoms. The gram atomic masses of sodium, nitrogen, and oxygen are 22.9898, 14.0067, and 15.9994 respectively. Therefore, the percentage by mass of nitrogen in the compound is: 100{14.0067/[14.0067 + 22.9898 + 3(15.9994)]} or about 16.4795, to the justified number of significant digits.
Calculate mass of 4.00 moles of sodium?
well take sodium, which the molar mass is 23.0 g/mol and multiply by # of moles (4) and you get 92g
Which is the mass of 8 moles of sodium chloride?
Sodium chloride has a molar mass of about 58.5 g/mol. So multiply 8 moles by molar mass to get about 468 grams.
How many moles of molecules are there in 250g of hydrogen nitrate?
Hydrogen nitrate has a mass of 63.01 g/mol. In order to find the number of moles you divide the grams by the molar mass. 250/63.01 = 3.96 mol.
What is the mass of 1 mol of NaNO3?
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
What is the percentage by mass of nitrate in sodium nitrate?
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
Which weighs more 5moles of sodium nitrate or 4 moles of carbon dioxide?
I'm assuming that your instructor is using the terms mass and weight interchangeably. First determine the molar mass of each compound by multiplying the molar mass of each element by its subscript and add them together. The molar mass of an element is its atomic weight on the periodic table in grams. Molar mass of sodium nitrate (NaNO3) (1 atom Na x 22.98970g/mol Na) + (1 atom N x 14.0067g/mol N) + (3 atoms O x 15.9994g/mol O) = 84.9946g/mol NaNO3 Molar mass of carbon dioxide (CO2) (1 atom C x 12.0107g/mol C) + (2 atoms O x 15.9994g/mol O) = 44.0095g/mol CO2 Now multiply the molar mass for each compound by the number given and compare the results to determine which has the greater mass. 5mol NaNO3 x 84.9946g/mol NaNO3 = 424.973g NaNO3 4 mol CO2 x 44.0095g/mol CO2 = 176.038 So 5 moles of sodium nitrate has a mass greater than 4 moles of carbon dioxide.
What is the molecular mass of sodium nitrate?
The molecular mass of sodium nitrate is 84,9947.
How do you Calculate how much sodium nitrate is needed to give 126g of nitric acid by reaction?
The reaction would be NaNO3 (s) + H2SO4 (l) -> HNO3 (l) + NaHSO4 (s)Moles ratio 1 : 1 : 1 : 1Moles of nitric acid:= Mass / Mr= 126g / HNO3= 126g / (1) + (14) + (16 x 3)= 126g / (1) + (14) + (48)= 126g / 63= 2 molsodium nitrate:nitric acidMoles ratio 1 : 1Actual moles in reaction 2 : 2Mass of sodium nitrate:= Moles x Mr= 2 x NaNO3= 2 x (23) + (14) + (16 x 3)= 2 x (23) + (14) + (48)= 2 x 85= 170gMass of sodium nitrate needed = 170g
What is the mass in grams of 0.254 mol sodium nitrate?
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
How many moles of nitrogen N are in 60.0 g of nitrous oxide N2O?
For this you need the atomic (molecular) mass of NaNO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaNO3=85.0 grams60.1 grams NaNO3 / (85.0 grams) = .707 moles NaNO3
How many atoms of sodium are present in Moles of sodium and what is the mass of it?
1. Three moles of sodium contain 18,06642387.1023 atoms. 2. The mass of three moles of sodium is 68,97 grams.
How many atoms of sodium are present in 3 moles of sodium and what is the mass of it?
1. Three moles of sodium contain 18,06642387.1023 atoms. 2. The mass of three moles of sodium is 68,97 grams.
Could you convert .10 mol of NaNO3 to grams?
#moles = mass/molar mass mass = #moles*molar mass mass = .10 moles*(atomic weight of na+atomic weight of N+3(atomic weight of oxygen)
What is the percentage by mass of nitrogen in sodium nitrate?
The formula for sodium nitrate is NaNO3, showing that each formula unit contains one sodium atom, one nitrogen atom, and three oxygen atoms. The gram atomic masses of sodium, nitrogen, and oxygen are 22.9898, 14.0067, and 15.9994 respectively. Therefore, the percentage by mass of nitrogen in the compound is: 100{14.0067/[14.0067 + 22.9898 + 3(15.9994)]} or about 16.4795, to the justified number of significant digits.
How much sodium nitrate is produced when 1.7G of silver nitrate and 1.5G of sodium iodide is reacted?
The equation for the reaction between silver nitrate and sodium iodide is AgNO3 + NaI -> AgBr + NaNO3. The gram formula masses are 169.87 for silver nitrate, 149.89 for sodium iodide, and 84.99 for sodium nitrate. Therefore, 1.7 g of silver nitrate constitutes 1.7/169.87 or 0.010 formula mass of silver nitrate and 1.5 g of sodium iodide constitutes 1.5/149.89 or 0.010 mole of sodium iodide, to the justified number of significant digits. The reaction equation shows that the number of formula unit masses of each reactant and product are the same, so that there will be 0.85 g of sodium nitrate produced, to the justified number of significant digits.
