To make a 3 M solution of barium chloride (BaCl₂), first calculate the required amount of BaCl₂ by using the formula: mass = molarity × volume × molar mass. The molar mass of barium chloride is approximately 208.23 g/mol. For example, to prepare 1 liter of a 3 M solution, dissolve 624.69 grams of barium chloride in enough water to make the total volume 1 liter. Always ensure to wear appropriate safety gear and follow proper lab protocols when handling chemicals.
There are 2. The chemical formula tells you that there is 1 barium and 2 chlorine.
This equation is 3 BaCl2 + 2 Ag3PO4 -> Ba3(PO4)2 + 6 AgCl.
The balanced chemical equation for this reaction is BaCl2 + Al2(SO4)3 -> 2AlCl3 + 3BaSO4. In this reaction, a double displacement reaction occurs where the cations and anions of the reactants switch partners to form the products. Barium sulphate, which is insoluble, precipitates out of the solution.
Al2(SO4)3 + 6KOH --> 2 [Al(OH)3](s) + 3K2SO4or in excess hydroxide:Al2(SO4)3 + 8KOH --> 2K+ + 2 [Al(OH)4]-(aq) + 3K2SO4
Correct. Your products will be barium nitrate, which is water soluble (all nitrates are soluble) and silver chloride, which is one of the few insoluble chlorides. There are three equations you can write for this reaction: 1. Normal balanced chemical equation: BaCl2 + 2AgNO3 --> Ba(NO3)2 + 2AgCl 2. Full ionic equation: Ba+2 + 2Cl- + 2Ag+ + NO3- --> Ba+2 + 2NO3- + 2Ag+ + 2Cl- 3. Net ionic equation: Ag+(aq) + Cl-(aq) --> AgCl(s)
There are 2. The chemical formula tells you that there is 1 barium and 2 chlorine.
This equation is 3 BaCl2 + Al2(SO4)3 -> 3 Ba(SO4) + 2 AlCl3.
The balanced chemical equation for Barium chloride plus Aluminium sulphate gives Barium sulphate Aluminium chloride is represented as .3BaCl2(aq) + Al2(SO4)3(aq) --> 3BaSO4(ppt) + 2AlCl3(aq).The ppt formed are white in color.
This equation is 3 BaCl2 + 2 Ag3PO4 -> Ba3(PO4)2 + 6 AgCl.
1. Put the mixture of powders in a beaker and add water. 2. Stir vigorously. Sodium chloride is dissolved, barium sulfate not. 3. Filter to separate sodium chloride solution (passes the filter) from barium sulfate as a solid on the filter.
This equation is: 3 BaCl2 (aq) + Al2(SO4)3 (aq) = 3 BaSO4 (s) + 2 AlCl3 (aq).
A mole of barium chloride, BaCl2, contains 6.022 x 10^23 formula units, and each formula unit contains 1 barium atom (Ba) and 2 chloride atoms (Cl). So, there are 6.022 x 10^23 barium atoms and 2 x 6.022 x 10^23 chloride atoms in a mole of barium chloride.
It would produce Barium oxide (BaO) and Chlorine gas (Cl2) BaCl2 + O2 --> BaO + Cl2
The balanced chemical equation for this reaction is BaCl2 + Al2(SO4)3 -> 2AlCl3 + 3BaSO4. In this reaction, a double displacement reaction occurs where the cations and anions of the reactants switch partners to form the products. Barium sulphate, which is insoluble, precipitates out of the solution.
Barium has a charge of -2 and Oxygen has a charge of +2. Oxygen is nature appears as O2, therefore Barium + Oxygen = barium oxide; Ba + O2 = BaO; 2Ba + O2 = 2BaO
Al2(SO4)3 + 6KOH --> 2 [Al(OH)3](s) + 3K2SO4or in excess hydroxide:Al2(SO4)3 + 8KOH --> 2K+ + 2 [Al(OH)4]-(aq) + 3K2SO4
Correct. Your products will be barium nitrate, which is water soluble (all nitrates are soluble) and silver chloride, which is one of the few insoluble chlorides. There are three equations you can write for this reaction: 1. Normal balanced chemical equation: BaCl2 + 2AgNO3 --> Ba(NO3)2 + 2AgCl 2. Full ionic equation: Ba+2 + 2Cl- + 2Ag+ + NO3- --> Ba+2 + 2NO3- + 2Ag+ + 2Cl- 3. Net ionic equation: Ag+(aq) + Cl-(aq) --> AgCl(s)