First, you want to figure out how many grams 1 mol of barium is. Look at the Periodic Table of elements. (www.webelements.com) Barium is Ba. Scroll down a little bit after clicking and you will see atomic weight. It is 137g.
Now get the atomic weight for chlorine. It is Cl. The atomic weight is 35.5g.
Just add them together. 137+35.5 = 172.5g = 1 mol of barium chloride.
For 188g, just divide it by 172.5g. Which is 1.09 moles.
Since both barium chloride and barium sulfate contain one mole of barium atoms pert mole of compound, the moles of barium sulfate will be the same, 0.100, when barium has the limiting concentration in the production of the sulfate.
I assume you mean 0.5 molar, but I'll go with what I'm given. If you have 500 molar of barium chloride, you will have 1000 molar of chloride ions as there are 2 chlorides for every 1 molecule of barium chloride. Having the 100ml there is irrelevant as you are talking about concentration and didn't ask for moles.
Since barium sulfate and barium chloride have a 1:1 molar ratio, you would need the same amount of barium chloride as barium sulfate, so 100 grams.
To find the number of moles of chloride ions in aluminum chloride, you first need to convert 0.2520g of aluminum chloride to moles. Then, since there are three chloride ions per one aluminum chloride molecule, you would multiply the number of moles of aluminum chloride by 3 to find the moles of chloride ions.
The compound formed between barium and chlorine would be barium chloride (BaCl2). In this compound, barium has a +2 charge and chlorine has a -1 charge, so two chlorine atoms are needed to balance the charge of one barium atom.
Since both barium chloride and barium sulfate contain one mole of barium atoms pert mole of compound, the moles of barium sulfate will be the same, 0.100, when barium has the limiting concentration in the production of the sulfate.
I assume you mean 0.5 molar, but I'll go with what I'm given. If you have 500 molar of barium chloride, you will have 1000 molar of chloride ions as there are 2 chlorides for every 1 molecule of barium chloride. Having the 100ml there is irrelevant as you are talking about concentration and didn't ask for moles.
BaCl would not exist. BaCl2 would be Barium Chloride, though.
To prepare a 2 m solution of barium chloride (BaCl₂), you need to know the molarity (moles per liter) and the molecular weight of BaCl₂, which is approximately 208.23 g/mol. For a 2 m solution, you need 2 moles per kilogram of solvent. Since 20 mL is 0.02 kg, you would need 0.04 moles of BaCl₂ (2 moles/kg × 0.02 kg). Multiplying 0.04 moles by the molar mass (208.23 g/mol) gives approximately 8.33 grams of barium chloride.
Since barium sulfate and barium chloride have a 1:1 molar ratio, you would need the same amount of barium chloride as barium sulfate, so 100 grams.
The compound formed from Ba and Cl would be named barium chloride.
The empirical formula for a substance containing 65.95% barium and 34.05% chlorine would be BaCl2 (barium chloride). This is determined by converting the mass percentage of each element to moles, finding the ratio of moles of each element, and simplifying to the simplest whole-number ratio.
When silver nitrate is added to barium chloride, a white precipitate of silver chloride forms because silver chloride is insoluble in water. This occurs due to a double displacement reaction where the silver ions from silver nitrate react with the chloride ions from barium chloride to form silver chloride. The remaining solution would contain barium nitrate as the other product of the reaction.
The valency of a barium ion is +2. Barium chloride (BaCl2) consists of a barium ion (Ba^2+) and two chloride ions (Cl^-), which gives the overall compound a neutral charge.
Nothing - barium chloride is soluble. You can however precipitate either the barium (e.g. with sodium sulphate, giving barium sulpate, or the chloride, e.g. with silver nitrate giving silver chloride precipitate.
The molar mass of barium (Ba) is approximately 137.3 g/mol. So, the mass of 3 moles of barium would be 3 moles x 137.3 g/mol = 411.9 grams.
Sulfuric acid cannot be used in place of hydrochloric acid in the barium chloride test because sulfuric acid would react with barium chloride to form insoluble barium sulfate. This would interfere with the precipitation reaction used to detect the presence of sulfates in the sample. Hydrochloric acid is preferred because it does not interfere with this reaction.