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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.
Yes, the molar mass of anhydrous sodium sulfate is 142,035 grams.
The mass is 234,8 g.
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
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 -------------------------------
One liter of Benedict's solution contains 173 grams sodium citrate, 100 grams sodium carbonate, and 17.3 grams cupric sulfate pentahydrate.
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.
Yes, the molar mass of anhydrous sodium sulfate is 142,035 grams.
The mass is 234,8 g.
How many grams of Na+ are contained in 25 g of sodium sulfate (Na2SO4)?
3.23
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
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 -------------------------------
Sodium sulfate is not prepared from hydrogen chloride.
Sodium Sulfate is Na2SO4 and the molecular weight is 142. So 0.15 moles would have a mass of 0.15 x142 = 21.3g.
4 milliequivalents of sodium chloride solution is a solution having 0,2338 g in 1 L.
4.5 grams