Yes, the molar mass of anhydrous sodium sulfate is 142,035 grams.
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
To determine the grams of sodium sulfate needed, you first need to specify the molarity (M) of the sodium sulfate solution. Once you have the molarity, you can use the formula: grams = molarity (M) x volume (L) x molar mass (g/mol). This will give you the amount of sodium sulfate in grams needed to make the solution.
Amount of sodium sulfate required = 0.683 x 350/100 = 0.239The formula mass of sodium sulfate, Na2SO4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 Therefore mass of sodium sulfate required = 0.239 x 142.1 = 34.0g Approximately 34 grams of sodium sulfate would be needed.
To solve this stoichiometry problem, first calculate the number of moles of sodium hydroxide (NaOH) present in 200 grams. Then, using the balanced equation, determine the moles of sodium sulfate (Na2SO4) that will be formed. Finally, convert the moles of Na2SO4 to grams using the molar mass of sodium sulfate.
The molecular weight of sodium sulfate (Na2SO4) is approximately 142.04 grams per mole.
There are 2.42 grams of sodium in 10 grams of sodium sulfate (Na2SO4). This is calculated based on the molecular weight of sodium sulfate and the molar ratio of sodium to the sulfate ion in the compound.
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
To determine the grams of sodium sulfate needed, you first need to specify the molarity (M) of the sodium sulfate solution. Once you have the molarity, you can use the formula: grams = molarity (M) x volume (L) x molar mass (g/mol). This will give you the amount of sodium sulfate in grams needed to make the solution.
Amount of sodium sulfate required = 0.683 x 350/100 = 0.239The formula mass of sodium sulfate, Na2SO4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 Therefore mass of sodium sulfate required = 0.239 x 142.1 = 34.0g Approximately 34 grams of sodium sulfate would be needed.
To solve this stoichiometry problem, first calculate the number of moles of sodium hydroxide (NaOH) present in 200 grams. Then, using the balanced equation, determine the moles of sodium sulfate (Na2SO4) that will be formed. Finally, convert the moles of Na2SO4 to grams using the molar mass of sodium sulfate.
To find the mass of sodium sulfate, we need to know the molar mass of Na2SO4, which is about 142.04 g/mol. Multiply the number of moles by the molar mass to get the mass: 0.150 mol * 142.04 g/mol = 21.31 grams. So, the mass of 0.150 mol of sodium sulfate is 21.31 grams.
Sodium sulfate is not prepared from hydrogen chloride.
Benedict's solution typically contains copper(II) sulfate, sodium citrate, and sodium carbonate. The exact composition can vary, but a common formulation is 17.3 g copper(II) sulfate, 100 g sodium citrate, and 173 g sodium carbonate per liter of water.
The molecular weight of sodium sulfate (Na2SO4) is approximately 142.04 grams per mole.
3.6 moles N2SO4 (142.05 grams/1 mole Na2SO4) = 511.38 grams Na2SO4 ==================( you do significant figures )
The gram formula mass of anhydrous sodium sulfate is 142.04. Therefore, 43 gram formula units* of it has a mass of 43(142.04)6.1 X 103 grams, to the justified number of significant digits. _____________________________ For ionic compounds such as sodium sulfate, the term "formula unit" is preferred to "mole".
To find the mass of Na+ in sodium sulfate (Na2SO4), we need to consider the molar ratios of Na+ in the compound. In Na2SO4, there are 2 Na+ ions for every 1 Na2SO4 unit. The molar mass of Na2SO4 is 142 g/mol, so in 25 g of Na2SO4, there are about 8.8 g of Na+.