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9 moles of bromine contain 54,2.10e23 molecules.
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.
2,9 moles of bromine is equivalent to 463,4432 g.
1,012 mole of bromine for the diatomic molecule.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
9 moles of bromine contain 54,2.10e23 molecules.
First convert the volume of the Br2 into grams by using:D=M/VSo we are given that volume=16.0 ml and density=3.12g/ml.M=D*VM=(3.12g/ml)*(16.0ml)=49.92 gThen we use #moles of a substance=#grams present/Formula weight(# of grams of Br2 in 1 mol of Br2)The Formula weight(molar mass) of Br2=2*(79.9 g/mol)=159.80 g/mol Br2#moles of Br2=49.92g/159.80g/mol Br2=.312 moles of Br2 present.
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.
Amount of Br2 = mass of sample / molar mass = 160 / 2(79.9) = 1.00mol
44.0 grams Br2 ? 44.0 grams Br2 (1 mole Br2/159.8 grams)(6.022 X 10^23/1 mole Br2)(1 mole Br2 atoms/6.022 X 10^23) = 0.275 moles of Br2 atoms
1 mole Br2 = 159.808g Br2 = 6.022 x 1023 molecules Br2 4.89 x 1020 molecules Br2 x 1mol Br2/6.022 x 1023 molecules Br2 x 159.808g Br2/mol Br2 = 0.130g Br2
0.275
2,9 moles of bromine is equivalent to 463,4432 g.
1,012 mole of bromine for the diatomic molecule.
The complete decomposition reaction is as follows:2 BrF3 → Br2 + 3 F2 , so 2 moles BrF3 will give 1 mole Br2 , hence 0.248 mole gives 0.124 mole Br2
One mole of Br2 has 6.023 x 1023 bromine molecules or 2 x 6.023 x 1023 bromine atoms.
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