Sodium sulfate powder is white.
its golden yellow
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
Na2SO4 + H2O = Na2SO4*10H2O > insoluble and foms clumbs what can be phisically separated from the solution
How many grams of Na+ are contained in 25 g of sodium sulfate (Na2SO4)?
It does. It's Na2SO4.
na2so4= 142.05 g/mole, use dimensional analysis & set up your problem 13.64g
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=============
25.0 grams Na2SO4 ( only way this compound is possible ) 25.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams) = 0.176 moles Na2SO4 -----------------------------
Na2SO4(s) --> 2Na+(aq) + SO4-2(aq)
Both NaOH and H2SO4 have no colour, even not after complete reaction (Na2SO4 and H2O are also colourless)
Because Na2SO4 is an ionic compound and so does not have actual molecules.
H2SO4 + 2NaOH ------------> Na2SO4 + 2H2O H2SO4 + 2NaOH ------------> Na2SO4 + 2H2O H2SO4 + 2NaOH ------------> Na2SO4 + 2H2O
Na2SO4 and NaCl
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 -------------------------------
inorganic
Na2SO4
8.20
Nothing. NaHSO4 + NaCl ---> HCl + Na2SO4