3 years ago

How many bromine atoms are present in 39.4 g of CH2Br2

1 answer:

7 0

37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2

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

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