How many grams are present in 5.6 litres of CO2 at N.T.P
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Step-by-step explanation:
At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate. 0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.
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Answered by
1
Answer:
At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles. They will be obtained from 0.25 moles of calcium carbonate. 0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.
Step-by-step explanation:
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