0
When 0.5 moles of Na2SO4 is dissolved in water how many moles of sodium ions are present?
Wiki User
∙ 13y ago
Since the formula shows two sodium atoms in each formula unit of sodium sulfate and this compound normally completely ionizes in water solution, the number of sodium ions will be twice the number of moles of the salt; in this instance, 1.0 moles of sodium ions.
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
How many moles are present in 585 g of sodium chloride?
There are 10 moles present in 585 g of sodium chloride.
How many grams of sodium are in 0.820 moles of Na2SO4?
na2so4= 142.05 g/mole, use dimensional analysis & set up your problem 13.64g
How many MOLES of sodium nitrate are present in 2.85 grams of this compound?
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
How many grams of solid barium sulfate form when 22.6 mL of 0.160 M barium chloride reacts with 54.6 mL of 0.055 M sodium sulfate?
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced
What is the mass in grams of 0.150 mol of sodium sulfate?
Sodium Sulfate is Na2SO4 and the molecular weight is 142. So 0.15 moles would have a mass of 0.15 x142 = 21.3g.
How many Na2SO4 molecules are present in 1.125 moles?
1,125 moles of sodium sulfate contain 6,774908464125.10e23 molecules.
How many moles of solute are present in 1.5 liters of 1.33 M of Na2SO4?
The answer is approx. 2 moles (for anhydrous sodium sulfate).
How many moles of sodium in 5.25 mole of Na2SO4?
5,25 moles (in anhydrous sodium sulphate)
What is the concentration of Na2SO4 solution prepared by dissolving 5.35 g of Na2SO4 in sufficient water to give 330 ml of solution?
First.Get moles sodium sulfate.5.35 grams Na2SO4 (1 mole Na2SO4/142.05 grams)= 0.0377 moles Na2SO4-------------------------------------Second.Molarity = moles of solute/Liters of solution ( 330 mL = 0.33 Liters )Molarity = 0.0377 moles Na2SO4/0.33 Liters= 0.114 M Na2SO4=============
How many grams are in 3.6 moles of sodium sulfate?
3.6 moles N2SO4 (142.05 grams/1 mole Na2SO4) = 511.38 grams Na2SO4 ==================( you do significant figures )
How many moles are present in 585 g of sodium chloride?
There are 10 moles present in 585 g of sodium chloride.
When calculating the number of moles of solid Na2SO4 the factor used is?
The molar mass of anhydrous sodium sulfate is 142,04.
How many grams of sodium are in 0.820 moles of Na2SO4?
na2so4= 142.05 g/mole, use dimensional analysis & set up your problem 13.64g
How many MOLES of sodium nitrate are present in 2.85 grams of this compound?
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
How many grams of solid barium sulfate form when 22.6 mL of 0.160 M barium chloride reacts with 54.6 mL of 0.055 M sodium sulfate?
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced
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 moles of which reactant are in excess when 359.0 mL of 0.259 M sulfuric acid reacts with 0.510 L of 0.191 M sodium hydroxide to form water and aqueous sodium sulfate?
Balanced equation first. 2NaOH + H2SO4 -> Na2SO4 +2H2O Molarity = moles of solute/Liters of solution 0.259 M H2SO4 = moles H2SO4/0.359 liters = 0.09298 moles H2SO4 0.191 M NaOH = moles of NaOH/0.510 liters = 0.09741 moles NaOH ( Let us find limiting reactant ) 0.09298 moles H2SO4 (1 mole H2SO4/2 mole NaOH) = 0.04649 moles H2SO4-- Sulfuric acid limits and, 0.04648 moles of sodium hydroxide are in excess.
