barium chloride plus sodium sulphate yields barium sulphate plus sodium chloride
Yes, the word 'equation' is a noun, a word for a thing.
Domain is the independent variable in an equation. It is what you put "in" the equation to get the Range.
x2
The abstract noun forms of the verb to 'equate' are equation and the gerund, equating (a word for a process).The noun 'equation' is an abstract noun as a word for the process of considering one thing to be the same as another; a word for a concept.The noun 'equation' is a concrete noun as a word for a statement with two sets of numbers, letters, or symbols are equal; a word for a written or spoken statement.
Equation
This equation is BaCl2 (aq) + Na2SO4 (aq) -> 2 NaCl (aq) + BaSO4 (s).
BaCl2(aq) + Na2SO4 (aq) ------> BaSO4(s) + 2NaCl(aq)
Test for the anions,according to your word,i think you got the salts,so, use these reagents HCL,BACL2,Adding bacl2 and Hcl to Naso4-white ppt observed in soluble,but barium chloride will give no ppt...so probem solved
To balance the chemical equation between BaCl2 and Na2SO4, you need to ensure that the number of each type of atom is the same on both sides of the equation. The balanced equation is BaCl2 + Na2SO4 -> BaSO4 + 2NaCl.
Ba2+ + [2Cl- + 2Na+] + SO42---> BaSO4 + [2Cl- + 2Na+]Ba2++ SO42- --> BaSO4
first of all, you need to recognize that one mole of Na2SO4 is reacting with one mole of BaCL2. so find the moles of the NaSO4, then you automatically have the moles of the BaCL2. if you get the moles of the BaCL2, its easy to calculate the volume of it because you already have the MOLARITY. good luck
To find the molarity of the BaCl2 solution, first calculate the moles of Na2SO4 in the sample using its molar mass. Then, use the balanced chemical equation of the precipitation reaction to determine the moles of BaCl2 needed to react with the moles of Na2SO4. Finally, divide the moles of BaCl2 by the volume of the solution in liters (57.0 mL = 0.057 L) to find the molarity.
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
The chemical equation for the reaction of sodium sulfate with barium chloride is: Na2SO4 + BaCl2 → 2NaCl + BaSO4. This is a double displacement reaction where the sodium and barium ions switch partners to form sodium chloride and barium sulfate.
Ooh. Somebody doesn't like doing their chem homework. Luckily I just happened to do that exact same equation. 1 BaCl2 +1 Na2S --> 2 NaCl(Salt! :D) +1 BaS
When BaCl2 (barium chloride) is added to Na2SO4 (sodium sulfate), a precipitation reaction occurs, resulting in the formation of a white precipitate of barium sulfate (BaSO4). This is represented by the chemical equation: BaCl2 + Na2SO4 → BaSO4 + 2NaCl.
Barium Chloride + Sodium Sulfate --> Barium Sulfate + Sodium Chloride BaCl2 + Na2So4 --> BaSO4 + 2NaCl It's called a Double Displacement reaction because Barium(Ba2+) and Sodium(Na+) displaces each other from their original anions. It's also called a Precipitation reaction because a white precipitate is formed after the reaction due to Barium Sulfate(BaSO4) as it is insoluble.