Want this question answered?
Bromine is a non metal element. Mass number of it is 80.
The gram formula unit or molar mass for aluminum bromide is 533.38.* Therefore, 1.42 moles has a mass of 757.4 grams. The mass of 6 moles of bromine atoms is 479.42. Therefore, the mass fraction of bromine in aluminum bromide is 479.42/757.4 or 0.633, and the mass in grams of bromine required to form 1.42 moles of aluminum bromide is 0.633 X 757.4 or 479 grams, to the justified number of significant digits (limited by the precision given for the number of moles.) ___________________ *This is equal to the sum of (2 times the gram atomic mass of aluminum) and (6 times the gram atomic mass of bromine).
Bromine isotopes have different mass numbers because their atoms have different numbers of neutrons.
81Br37Cl=117.882 81Br35Cl= 115.885 79Br37Cl=115.884 79Br35Cl =113.887
The atomic mass of boron is 10.811 while that of bromine is 79.904. Thus the ratio of the mass of bromine to the total mass of boron tribromide must be: 3(79.904)/[3(79.904) + 10.811] = 0.9568. The required mass of BBr3 thus is 11.8/0.9568 = 12.3 g, to the justified number of significant digits.
Amount of Br2 = mass of sample / molar mass = 160 / 2(79.9) = 1.00mol
Bromine (Br) has a molar mass of 79.904 amu (atomic mass units), which is extremely close to 80. Bromine is diatomic so when two bromine molecules are put together to create a diatomic gas, the molar masses of each bromine add to get a combined molar mass of 160 amu.
Bromine is a non metal element. Mass number of it is 80.
Bromine has an approximate atomic mass of 79.904.
Bromine is a metal element. Atomic mass of it is 79.90.
567.1 mass is the equal to 56789.0 intergram. XD
Bromine is a non meta element. Atomic mass of it is 80.
Bromine is a non metal element. Atomic mass of it is 80.
Mol = mass/Ar (relative atomic mass)For Bromine Ar is 79.9 g/mol.Mol= 0.476/79.9 = 5.96x10-3
The gram formula unit or molar mass for aluminum bromide is 533.38.* Therefore, 1.42 moles has a mass of 757.4 grams. The mass of 6 moles of bromine atoms is 479.42. Therefore, the mass fraction of bromine in aluminum bromide is 479.42/757.4 or 0.633, and the mass in grams of bromine required to form 1.42 moles of aluminum bromide is 0.633 X 757.4 or 479 grams, to the justified number of significant digits (limited by the precision given for the number of moles.) ___________________ *This is equal to the sum of (2 times the gram atomic mass of aluminum) and (6 times the gram atomic mass of bromine).
Density is equal to the mass (in g) divided by the volume (in mL). So 25.0 m/L divided by 77.58 g is your answer.
52