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Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃(Atomic mass Ba=137) will be
Options
(a) 2.24 L
(b) 4.96 L
(c) 1.12L
(d) 0.84L
Correct Answer:
1.12L
Explanation:
BaCO₃ → BaO+ CO₂
Atomic Mass of Ba=137
Atomic Mass of C=12
Atomic Mass of O=16
Molecular Mass of BaCO₃ = 137 + 12 + (16*3) = 197
197 gm of BaCO₃ released carbondioxide = 22.4 litre at STP
1 gm of BaCO₃ released carbondioxide = 22.4/197 litre
9.85 gm of BaCO₃ released carbondioxide = 22.4/197 x 9.85 = 1.12 litre
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The variation of the boiling points of the hydrogen halides is in the order
- Which of the following depends on the path followed
- The equivalent weight of H₃PO₂, when it disproportionates into PH₃ and H₃PO₃ is
- For making distinction between 2-pentanone and 3-pentanone the reagent to be
- By what factor does the average velocity of a gaseous molecule increase
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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