Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq).
An example is barium sulfate:BaCl2 + Na2SO4 = BaSO4 + 2 NaCl
Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq). The reaction results in the formation of insoluble barium sulfate (BaSO4).
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=============
To find the number of moles of Na2SO4 in 25.0 g of the compound, you need to convert the mass to moles. First, determine the molar mass of Na2SO4, then divide the given mass by the molar mass to obtain the number of moles.
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 nitrate.
Precipitation of barium sulfate:BaCl2 + Na2SO4 = BaSO4(s) + 2 NaCl
An example is barium sulfate:BaCl2 + Na2SO4 = BaSO4 + 2 NaCl
Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq). The reaction results in the formation of insoluble barium sulfate (BaSO4).
The reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) results in the formation of sodium sulfate (Na2SO4) and water (H2O). The balanced chemical equation is: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
The reaction between Na2SO4 and HCl is a chemical change because it results in the formation of new substances (NaCl and H2SO4) with different chemical compositions compared to the reactants.
The reaction between BaCl2 and Na2SO4 is a double displacement reaction, also known as a precipitation reaction. This means that the cations and anions of the two compounds switch partners to form two new compounds, and one of the products, BaSO4, is insoluble and precipitates out of solution.
The carbon mineral bondinh
To find the molality, we first calculate the moles of Na2SO4: 10.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 0.0705 moles Na2SO4. Then, molality is calculated as moles of solute (Na2SO4) / kilograms of solvent (water): 0.0705 mol / 1.000 kg = 0.0705 mol/kg, which is the molality of the 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=============
precipitation
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+.