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
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
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 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
Br2 + 3NaHSO3 = 2NaBr + NaHSO4 + H2O + 2SO2
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
9 moles of bromine contain 54,2.10e23 molecules.
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
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.
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
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
The mass of 0.030 moles of Br2 is 4.79424 grams, properly rounded to 4.8 grams.
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.
2LiBr(aq) + Cl2(g) = 2LiCl(aq) + Br2(l) will result in .167 moles of lithium chloride.
First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr