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How many moles of Al are needed to form 2.43 mol of Al2Br6?
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.
What is the oxidation number of Br in Al2Br6?
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.
What is the oxidation number of Al in Al2Br6?
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.
Aluminum will react with bromine to form aluminum bromide Al2Br6 What mass of bromine g is needed to form 1.42 mol of Al2Br6?
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.
What is the empirical formula for Al2Br6?
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.
