FeBr3 (Iron III Bromide) has three moles of bromide for every mole of iron. FeBr2 (Iron II Bromide) has two moles of bromide ion per mole of Iron.
To find the number of moles in 1.2 kg of calcium bromide, you first need to determine the molar mass of calcium bromide (CaBr2), which is approximately 199.89 g/mol. Then convert the mass of 1.2 kg to grams (1200 g). Finally, divide the mass in grams by the molar mass to find the number of moles. In this case, 1200g / 199.89g/mol ≈ 6 moles of calcium bromide.
To calculate the grams of lithium bromide present in the solution, you would first determine the moles of lithium bromide using the formula: moles = Molarity x Volume (L). Once you have the moles, you can convert it to grams using the molar mass of lithium bromide (86.85 g/mol).
That is a lot of calcium bromide we are dealing with. The formula mass of calcium bromide, CaBr2 is 40.1 + 2(79.9) = 199.9.Amount of CaBr2 = (7.4 x 1000)/199.9 = 32.0mol There are 32 moles of calcium bromide in a 7.4kg pure sample.
The balanced chemical equation for the reaction is: 2LiBr + Cl2 -> 2LiCl + Br2 Since the ratio of lithium bromide to lithium chloride is 1:1, 0.046 mol of lithium bromide will produce 0.046 mol of lithium chloride.
The equation of the reaction is BaBr2 + 2 AgNO3 -> 2 AgBr + Ba(NO3)2. Therefore, exactly as many bromide ions from barium bromide must be supplied to precipitate any particular number of silver ion from silver nitrate. From the definition of molarity, 100 ml of 52 M solution contains 5.2 moles (preferably called "gram formula units") of silver nitrate. The gram formula unit mass of silver nitrate is 169.87, and each gram formula mass contains equal numbers of silver and of nitrate ions. Therefore, 5.2 gram elemental masses of bromide ions will be required for the precipitation. This amount of bromide ions can be supplied by 5.2/2* or 2.6 gram formula masses of barium bromide, and the gram formula unit mass of barium bromide is 297.14. Multiplying this number by 2.6 shows that 7.7 X 102 grams of barium bromide, to the justified number of significant digits, will be needed.
If you think to a simple binary bromide as NaBr: 166,57.10e23 atoms.
In iron bromide FeBr2, there is 1 iron atom present.
Ther answer is none! ammonium bromide is made from hydrogen bromide and ammonia NH3 + HBr = NH4Br i mole of each makes 1mole of ammonium salt.
There are 1.35 moles of MgBr2 in 1 L of solution, which corresponds to 2 moles of bromide ions. Therefore, in 750.0 mL of 1.35 M MgBr2 solution, there will be 1.0125 moles of bromide ions.
The molecular weight of Calcium Bromide is extremely close to 200. So 1200/200 = 6 moles present.
To find the number of moles in 1.2 kg of calcium bromide, you first need to determine the molar mass of calcium bromide (CaBr2), which is approximately 199.89 g/mol. Then convert the mass of 1.2 kg to grams (1200 g). Finally, divide the mass in grams by the molar mass to find the number of moles. In this case, 1200g / 199.89g/mol ≈ 6 moles of calcium bromide.
To calculate the grams of lithium bromide present in the solution, you would first determine the moles of lithium bromide using the formula: moles = Molarity x Volume (L). Once you have the moles, you can convert it to grams using the molar mass of lithium bromide (86.85 g/mol).
That is a lot of calcium bromide we are dealing with. The formula mass of calcium bromide, CaBr2 is 40.1 + 2(79.9) = 199.9.Amount of CaBr2 = (7.4 x 1000)/199.9 = 32.0mol There are 32 moles of calcium bromide in a 7.4kg pure sample.
To calculate the mass of calcium bromide needed, you would first find the number of moles needed using the equation moles = Molarity * Volume (in liters). Then, you would use the molar mass of calcium bromide to convert moles to grams. The molar mass of calcium bromide is 199.89 g/mol.
To determine how many liters of a 4M lithium bromide (LiBr) solution can be made from 100 grams of LiBr, we first need to calculate the number of moles in 100 grams. The molar mass of lithium bromide is approximately 86.84 g/mol, so 100 grams corresponds to about 1.15 moles. A 4M solution contains 4 moles of solute per liter, thus 1.15 moles can produce approximately 0.29 liters (1.15 moles ÷ 4 moles/L). Therefore, 100 grams of lithium bromide can make about 0.29 liters of a 4M solution.
To find the number of moles in 245g of potassium bromide, first calculate the molar mass of KBr by adding the atomic masses of potassium (39.10 g/mol) and bromine (79.90 g/mol). The molar mass of KBr is 119.00 g/mol. Then, divide the given mass by the molar mass: 245g / 119.00 g/mol = 2.06 moles of potassium bromide.
The balanced chemical equation for the reaction is: 2LiBr + Cl2 -> 2LiCl + Br2 Since the ratio of lithium bromide to lithium chloride is 1:1, 0.046 mol of lithium bromide will produce 0.046 mol of lithium chloride.