1 mole of Al2Br6 contains 2 moles of Al, so to form 2.43 mol of Al2Br6, you will need 2.43 x 2 = 4.86 mol of Al.
The oxidation number of Br in Al2Br6 is -1. This is because the overall charge of the compound must be zero, and since we have two Br atoms each at -1 oxidation state, it balances out with the +3 oxidation state of Al.
The oxidation number of Al in Al2Br6 is +3. Each bromine atom has an oxidation number of -1, and since the compound is neutral, the sum of the oxidation numbers must equal zero. Hence, each Al atom must have an oxidation number of +3 to balance the -6 from the bromine atoms.
To find the mass of bromine needed to form 1.42 mol of Al2Br6, first calculate the molar mass of Al2Br6 which is 531.55 g/mol. Since the molar ratio of aluminum to bromine in Al2Br6 is 2:6, for 1.42 mol of Al2Br6, you would need (1.42 mol * 6 mol of bromine/mol of Al2Br6) = 8.52 mol of bromine. Finally, multiply the molar mass of bromine (79.90 g/mol) by the number of moles needed (8.52 mol) to get the mass of bromine required, which is 680 g.
The empirical formula for Al2Br6 is AlBr3. This is because the subscripts must be reduced to the simplest whole number ratio, which in this case is 1:3 for aluminum to bromine atoms.
The molecular name for AL2Br6 is aluminum tribromide.
1 mole of Al2Br6 contains 2 moles of Al, so to form 2.43 mol of Al2Br6, you will need 2.43 x 2 = 4.86 mol of Al.
The oxidation number of Br in Al2Br6 is -1. This is because the overall charge of the compound must be zero, and since we have two Br atoms each at -1 oxidation state, it balances out with the +3 oxidation state of Al.
The oxidation number of Al in Al2Br6 is +3. Each bromine atom has an oxidation number of -1, and since the compound is neutral, the sum of the oxidation numbers must equal zero. Hence, each Al atom must have an oxidation number of +3 to balance the -6 from the bromine atoms.
The formula for the ionic compound formed between aluminum (Al) and bromine (Br) is AlBr3. Aluminum is a metal with a charge of +3, while bromine is a nonmetal with a charge of -1. Therefore, three bromine atoms are needed to balance the charge of one aluminum atom.
To find the mass of bromine needed to form 1.42 mol of Al2Br6, first calculate the molar mass of Al2Br6 which is 531.55 g/mol. Since the molar ratio of aluminum to bromine in Al2Br6 is 2:6, for 1.42 mol of Al2Br6, you would need (1.42 mol * 6 mol of bromine/mol of Al2Br6) = 8.52 mol of bromine. Finally, multiply the molar mass of bromine (79.90 g/mol) by the number of moles needed (8.52 mol) to get the mass of bromine required, which is 680 g.
The empirical formula for Al2Br6 is AlBr3. This is because the subscripts must be reduced to the simplest whole number ratio, which in this case is 1:3 for aluminum to bromine atoms.
What is the chemical formula for Aluminum bromide?AlBr3(s) Al3+(aq) + 3Br-(aq)
Lithium bromide, LiBr, is prepared by the treatment of lithium carbonate with hydrobromic acid
Aluminum bromide (AlBr3) is considered an ionic compound because it is composed of positively charged aluminum ions (Al3+) and negatively charged bromide ions (Br-). In the compound, aluminum donates three electrons to bromine in order to achieve a stable octet configuration and form a strong electrostatic attraction between the ions, resulting in an ionic bond.
Aluminum bromide is an ionic bond, formed by the transfer of electrons from aluminum to bromine to create positively charged ions (Al3+) and negatively charged ions (Br-).
It is a compound.