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
0.275
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
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
covalent because Br2 is just to Bromine atoms bonded together
10
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
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.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
There are two bromine atoms in Br2
Br2 is known as a diatomic molecule. It has 2 atoms, both of which are bromine atoms (Br).
One mole of Br2 has 6.023 x 1023 bromine molecules or 2 x 6.023 x 1023 bromine atoms.
0.275
9 moles of bromine contain 54,2.10e23 molecules.
2,9 moles of bromine is equivalent to 463,4432 g.
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
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