1 mol of bromine has 79.904g. 79.904g/mol x 350mol =
The atomic mass of hydrogen is 1u. Atomic mass of bromine is 80u. Therefore, the molar mass of hydrogen bromide is 1+80=81u.
The atomic mass of AgBr, which is a compound made up of silver (Ag) and bromine (Br) is the sum of the atomic masses of silver and bromine. Silver has an atomic mass of approximately 107.87 g/mol, while bromine has an atomic mass of approximately 79.904 g/mol. Therefore, the atomic mass of AgBr is approximately 107.87 + 79.904 = 187.774 g/mol.
The molar mass of tribromine oxide (Br3O) can be calculated by adding up the atomic masses of its constituent atoms. The molar mass of bromine (Br) is approximately 79.90 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol. Therefore, the molar mass of tribromine oxide is approximately 249.70 g/mol.
Yes, that is what is used. Bromine is I think ~51:49 ratio of Br79 and Br81. Therefore it is somewhere in between (79.9g/mol)
To find the grams of bromine (Br) in 595 g of calcium bromide (CaBr2), first determine the molar mass of CaBr2. The molar mass is approximately 40.08 g/mol for Ca and 79.90 g/mol for Br, giving a total of about 199.89 g/mol for CaBr2. Since there are two bromine atoms in each formula unit, the mass of bromine in CaBr2 is about 159.80 g (2 × 79.90 g). Thus, the mass of Br in 595 g of CaBr2 can be calculated as follows: (159.80 g Br / 199.89 g CaBr2) × 595 g CaBr2 ≈ 477.23 g of Br.
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).
The chemical formula of potassium bromide is KBr, showing that each formula unit contains equal numbers of potassium and bromine atoms. The gram atomic masses of potassium and bromine are 39.0983 and 70.904 respectively. Therefore, the mass fraction of bromine in KBr is 70.904/(70.904 + 39.0983) or about 0.644568. 50.0 g of potassium bromide therefore contains 32.2 g of bromine, to the justified number of significant digits.
The molar mass of CaBr2 (calcium bromide) is calculated by adding the atomic masses of calcium (Ca) and two bromine (Br) atoms. The atomic mass of calcium is approximately 40.08 g/mol and bromine is approximately 79.90 g/mol. Therefore, the molar mass of CaBr2 is 40.08 + (2 * 79.90) = 199.88 g/mol.
To determine the number of moles of bromine gas in 37.7 grams, you need to divide the given mass by the molar mass of bromine. The molar mass of bromine is approximately 79.904 g/mol. So, 37.7 grams of bromine is equal to 0.471 moles (37.7 g ÷ 79.904 g/mol).
Bromine has an atomic mass of 79.904 g/mol, while the molar mass of NaBr is 102.89 g/mol. Therefore, the percent composition of bromine in NaBr is (79.904 g/mol / 102.89 g/mol) x 100 = 77.68%.
The molar mass of PbBr2 is 367.008 g/mol
To find the number of moles in 0.476 grams of bromine, you first need to determine the molar mass of bromine, which is approximately 79.904 g/mol. Then, you can use the formula: moles = mass (g) / molar mass (g/mol). Therefore, in this case, 0.476 grams of bromine is equivalent to 0.006 moles.
Argon is 4 times as heavy as neon atoms. Neon has an atomic number of 10 and an atomic mass of about 20.18 g/mol, while argon has an atomic number of 18 and an atomic mass of about 39.95 g/mol.
The atomic mass of hydrogen is 1u. Atomic mass of bromine is 80u. Therefore, the molar mass of hydrogen bromide is 1+80=81u.
The atomic mass of AgBr, which is a compound made up of silver (Ag) and bromine (Br) is the sum of the atomic masses of silver and bromine. Silver has an atomic mass of approximately 107.87 g/mol, while bromine has an atomic mass of approximately 79.904 g/mol. Therefore, the atomic mass of AgBr is approximately 107.87 + 79.904 = 187.774 g/mol.
The molar mass of the compound CaBr2 is 199.9 grams per mole.
One molecule of bromine contains 2 atoms. Therefore, to find the number of grams in 602200000000000000000000 atoms, you would divide this number by Avogadro's number (6.022 x 10^23) to get the number of moles, and then convert to grams using the molar mass of bromine (79.9 g/mol).