joshmaran98 joshmaran98
  • 13-09-2017
  • Chemistry
contestada

How many bromine atoms are present in 39.4 g of CH2Br2

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mikamikaovf2g9 mikamikaovf2g9
  • 13-09-2017
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2 

4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms
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