Bromine exists as a diatomic gas. Thus, there are two moles of bromine atoms in 1 mole of bromine gas.
To convert atoms of bromine to moles of bromine, you would need to divide the number of atoms by Avogadro's number (6.022 x 10^23). So, to get moles of bromine, you would divide the number of atoms by Avogadro's number, not multiply.
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.
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.
Balanced equation. 2Fe + 3Br2 -> 2FeBr3 need limiting reactant; suspect iron 9.20 mole Fe (3 mole Br2/2 mole Fe) = 13.8 moles the stated iron moles are not nearly this much; iron limits and drives the reaction looks one to one, so................ 9.20 moles Fe (2 moles FeBr3/ 2mole Fe) = 9.20 moles FeBr3 produced
10,0 moles of bromine atoms contain 60,22140857.1023 atoms.Attention: valid for bromine atoms !.
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.
To convert atoms of bromine to moles of bromine, you would need to divide the number of atoms by Avogadro's number (6.022 x 10^23). So, to get moles of bromine, you would divide the number of atoms by Avogadro's number, not multiply.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
To find the number of moles of bromine atoms in a sample of (2.03 \times 10^{24}) atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Divide the number of atoms by Avogadro's number: [ \text{moles of Br} = \frac{2.03 \times 10^{24}}{6.022 \times 10^{23}} \approx 3.37 \text{ moles}. ] Therefore, the sample contains approximately 3.37 moles of bromine atoms.
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
The equation 2 P + 3 Br2 -> 2 PBr3 shows the presence of two phosphorus atoms and 6 bromine atoms on each side of the equation. Therefore, the atomic ratio of P to Br is 1:3; both integers are exact. The gram atomic masses of phosphorus and bromine are 30.9738 and 79.904 respectively. Therefore, the mass ratio of Br to P is (3*)(79.904)/30.9738 or 7.73919, to the justified number of significant digits. ______________________ *This integer is exact.
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
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
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.
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