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Assuming full decomposition, the volume of `CO_(2)` released at STP on heating 9.85 g of `BaCO_(3)` (At mass Ba = 137) will beA. 0.84 LB. 0.24 LC. 4.06 LD. 1.12 L |
Answer» Correct Answer - D `underset(197)(BaCO_(3))toBaO+underset(22.4L)(CO_(2))` `V_(CO_(2))` released `= (22.4)/(197)xx9.85 = 1.12L` |
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