Bromine exists as a diatomic gas. Thus, there are two moles of bromine atoms in 1 mole of bromine gas.
One mole of Br2 has 6.023 x 1023 bromine molecules or 2 x 6.023 x 1023 bromine atoms.
.0326 moles
1.54 (mol Br2) * 6.022*10+23 (molecule/mol Br2) * 2 (atoms Br/molecule Br2) =1.85*1024 atoms in 1.54 mole Br2
.467 mol of Bromine gas
The chemical equation is 2Al+3Br2 --> 2AlBr3 So find the number of moles of the aluminum and bromine. Moles of aluminum = 31 g Al/ 27 g/mol = 1.15 moles Moles of bromine = 97.5g/ 159.8 g/mol = .61 moles Then you find the limiting reagent by dividing the number of moles of each substance by their coefficient. 1.15 moles Al / 2 = .575 .61 moles Br / 3 = .203 That means that .203 *2 moles of AlBr3 was produced, because once the Br reactant runs out, the reaction stops and you cannot get anymore of the product. So .406 moles of AlBr3 multiplied by the molecular weight of AlBr3, 266.7 g/mol, yields 108.28 grams of AlBr3
10,0 moles of bromine atoms contain 60,22140857.1023 atoms.Attention: valid for bromine atoms !.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
One mole of Br2 has 6.023 x 1023 bromine molecules or 2 x 6.023 x 1023 bromine atoms.
If it is 1.54 moles of Br atoms then the answer is 9.274 X 1023 atoms.If it is 1.54 moles of Br2 molecules then the answer is 1.855 X 1024 atoms.
The atoms in the reacts are always present in the products. There is one mole of bromine per molecule and .196 moles of the molecule. Thus, there will be .196 mols of bromine present after the reaction.
Bromine at standard temperature has diatomic molecules, and by definition one mole of anything has Avogadro's Number of molecules. Therefore, 2.6 moles of bromine contain 2(exact) X 2.6 X 6.022 X 1023 or 3.1 X 1024 atoms, to the justified number of significant digits.
.0326 moles
The formula unit for strontium chloride is StCl2 and therefore contains 3 atoms, one of strontium and two of chlorine.
24.5 mL of a solution 1.0 M bromine contain 0,0245 moles.
1.54 (mol Br2) * 6.022*10+23 (molecule/mol Br2) * 2 (atoms Br/molecule Br2) =1.85*1024 atoms in 1.54 mole Br2
1 mol of Br2 contains 6.02 x 1023 molecules (avogadro constant). In each Br2 molecule there are two Br atoms. Thus, number of Br atoms = 2 x 2.71 x 6.02 x 1023 = 3.26 x 1024
This is equivalent to 0,36 moles.