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babunello
[35]
3 years ago
5
How many bromine atoms are present in 37.5 g of CH2Br2?
Chemistry
Answer
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1
answer:
ella
[17]
3 years ago
5
0
<span>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</span>
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