Aluminum Tribromide, it is tri, because there are three Bromine in it. 'Tri' means three.
AlBr3 is the formula for aluminum (III) bromide.
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
The charge on the bromide ion (Br-) in AlBr3 is -1. This is because aluminum (Al) has a charge of +3, so in order to balance the charges and create a neutral compound, each bromide ion must have a charge of -1.
Aluminum can form aluminum bromide (AlBr3) when it reacts with bromine. This compound is primarily used as a catalyst in organic synthesis reactions.
The formula for the compound formed between aluminum and bromine is AlBr3, where aluminum has a +3 charge and bromine has a -1 charge. The subscript 3 in the formula indicates that there are three bromine atoms for every one aluminum atom in the compound.
The chemical name for aluminum bromide is aluminum tribromide, with the chemical formula AlBr3.
AlBr3 is the formula for aluminum (III) bromide.
There is no such compound as BaBr3, there is BaBr2, barium bromide.
What is the chemical formula for Aluminum bromide?AlBr3(s) Al3+(aq) + 3Br-(aq)
Given the balanced equation2Al + 6HBr --> 2AlBr3 + 3H2In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr----------- 266.6g AlBr3 * 2 molecules AlBr3
what is the cation for Ca(ClO4)2
The molecular name for AL2Br6 is aluminum tribromide.
When AlBr3 ionizes, it will produce one Al3+ ion and three Br- ions.
AlBr3
sp3
The nonmetal in these compounds are oxygen in MgO, hydrogen in NaH, bromine in AlBr3, and sulfur in FeS.
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