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Andrew [12]
3 years ago
9

Please help me ASAP PLEASE​

Chemistry
1 answer:
vodomira [7]3 years ago
3 0
24 cm^3 CO2 / 1 * 1 dcm^3 / 1000 cm^3 * 1 mol / 24 dcm^3 = 0.001 mol of CO3 is required
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