Three (3)
Here is the equation
Al^3+ + 3Br^- = AlBr3
Two bromide ions can combine with one barium cation to form an ionic compound, because a barium cation has an electrical charge of +2, while a bromide anion has an electrical charge of -1.
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.
Aluminum chloride contains 3 chlorine atoms per molecular unit. Therefore, in 3 moles there are 3 times Avogadro's number of chloride ions = 1.807 X 1024.
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.
Molarity is mols of solute/volume of solution. We will do this in two parts, though you could use one conversion string. 0.300 M AlBr3 = mols/0.500L = 0.15 mols of AlBr3 (3mol Br/1 mol AlBr3 )( 6.022 X 10^23/1mol Br )(1mol Br/6.022 X 10^23 ) = 0.45 moles of Br
Two bromide ions can combine with one barium cation to form an ionic compound, because a barium cation has an electrical charge of +2, while a bromide anion has an electrical charge of -1.
Bromine has -1 charge and Lithium has +1 charge. Therefore,only one lithium ion is required to react with a bromine ion.
To determine how many moles of bromide are in iron (III) bromide (FeBr3), you can use the chemical formula FeBr3 to see that there are three moles of bromide ions for every mole of iron (III) bromide. So, the number of moles of bromide ions is equal to the number of moles of FeBr3.
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.
There would be 4.38 moles of fluoride ions in 1.46 moles of aluminum fluoride, as the formula for aluminum fluoride is AlF3 with three fluoride ions per molecule of aluminum fluoride.
One molecule of aluminum bromide contains one aluminum atom and three bromine atoms, totaling four atoms.
Aluminum is 3+ and chloride is 1-. So, you need 3 chloride ions to neutralize 1 aluminum ion.
One formula unit of MgBr2 has three ions; one Mg2+ ion and two Br- ions. One mole of MgBr2 formula units has one mole of Mg2+ ions and two moles of Br- ions, for a total of three moles of ions.
There are 3 ions in the formula Al2SO4 (2 aluminum and 1 sulfate), but this formula is wrong. The formula of aluminum sulfate is Al2(SO4)3 which contains 5 ions: 2 aluminum ions and 3 sulfate ions.
Three chlorine ions are required to bond with one aluminum ion in order to form the compound aluminum chloride. This results in a stable compound with a 1:3 ratio of aluminum to chlorine ions.
Three chlorine ions are required to bond with one aluminum ion to form the compound aluminum chloride. This is because aluminum has a 3+ charge and chlorine has a 1- charge, so the formula for aluminum chloride is AlCl3.
Two. covalent bonds in the ammonium ion NH4+ and ionic bond between the ammonium ion and the bromide ion, Br-