No it is not an acid.It is a neutral compound.
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
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
Br2 + 3NaHSO3 = 2NaBr + NaHSO4 + H2O + 2SO2
CH4 + Br2 = CH3Br + HBr
Br2 + 2NaI -> 2NaBr + I2
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
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
Br2 + 3NaHSO3 = 2NaBr + NaHSO4 + H2O + 2SO2
C6H6O + Br2
If you mean Br2, it is a compound.
There are two bromine atoms in Br2
3.387mL Br2
CH4 + Br2 = CH3Br + HBr
Br2 + Zn ----> ZnBr2
Bromine itself is not considered an acid. However, it can react with water to a small degree to form hydrobromic acid (HBr), a strong acid and hypobromous acid (HBrO) a weak acid. Br2 + H2O --> HBr + HBrO
Br2 + 2NaI -> 2NaBr + I2
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.