One mole of Br2 has 6.023 x 1023 bromine molecules or 2 x 6.023 x 1023 bromine atoms.
Bromine exists as a diatomic gas. Thus, there are two moles of bromine atoms in 1 mole of bromine gas.
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
To determine the number of moles in 2.60100 grams of bromine, we first need to calculate the molar mass of bromine, which is 79.904 g/mol. Then, we use the formula: moles = mass / molar mass. Substituting the values, we get moles = 2.60100 g / 79.904 g/mol = 0.03255 moles of bromine.
Avogadro's constant is 6.02*10^23. This number represents the number of representative particles (atoms, molecules, or formula units) in one mole. To solve your question, we simply multiply Avogadro's constant by the number of moles. 6.02*10^23 * 3.01 = 1.81*10^24
To determine the number of molecules in 120 grams of bromine gas, you first need to calculate the moles of bromine using its molar mass (molar mass of Br2 = 159.808 g/mol). Then, use Avogadro's number (6.022 x 10^23) to find the number of molecules in that many moles of bromine gas.
10,0 moles of bromine atoms contain 60,22140857.1023 atoms.Attention: valid for bromine atoms !.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
Bromine exists as a diatomic gas. Thus, there are two moles of bromine atoms in 1 mole of bromine gas.
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.
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
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
Bromine at standard temperature has diatomic molecules, and by definition one mole of anything has Avogadro's Number of molecules. Therefore, 2.6 moles of bromine contain 2(exact) X 2.6 X 6.022 X 1023 or 3.1 X 1024 atoms, to the justified number of significant digits.
The atoms in the reacts are always present in the products. There is one mole of bromine per molecule and .196 moles of the molecule. Thus, there will be .196 mols of bromine present after the reaction.
one mole of a substance is described as 6.02x1023 atoms of a substance so if one mole of bromide gas contains 6.02x1023 atoms then bromide gas will contain one mole. your question is a trick question as the gas is stated as containg one mole there fore it contains one mole of bromide atoms
There are 6.022x1023 atoms in a mole. You multiply 6.022x1023 by 8.68, which equals 52.20796x1023 atoms
To determine the number of moles in 2.60100 grams of bromine, we first need to calculate the molar mass of bromine, which is 79.904 g/mol. Then, we use the formula: moles = mass / molar mass. Substituting the values, we get moles = 2.60100 g / 79.904 g/mol = 0.03255 moles of bromine.
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