This is equivalent to 0,36 moles.
9 moles of bromine contain 54,2.10e23 molecules.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
10,0 moles of bromine atoms contain 60,22140857.1023 atoms.Attention: valid for bromine atoms !.
24.5 mL of a solution 1.0 M bromine contain 0,0245 moles.
2,9 moles of bromine is equivalent to 463,4432 g.
Since one mole is equal to 6.022x10^23, there are .36 moles in 2.17x10 representative particles of bromine. A mole is a measure used to make atomic calculations for density.
9 moles of bromine contain 54,2.10e23 molecules.
avagadros number
0,2 moles contain approx. 1,2.10e+23 atoms, molecules etc.
There are 1.5 x 10^24 representative particles in 2.5 moles, considering Avogadro's number (6.022 x 10^23) representative particles per mole.
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.
2,60x102 grams of bromine (Br) is equal to 1,627 moles Br2.
10,0 moles of bromine atoms contain 60,22140857.1023 atoms.Attention: valid for bromine atoms !.
24.5 mL of a solution 1.0 M bromine contain 0,0245 moles.
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
The complete decomposition reaction is as follows:2 BrF3 → Br2 + 3 F2 , so 2 moles BrF3 will give 1 mole Br2 , hence 0.248 mole gives 0.124 mole Br2
Bromine exists as a diatomic gas. Thus, there are two moles of bromine atoms in 1 mole of bromine gas.