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The number of molecules of $CO_2$ liberated by the complete combustion of $0.1$ grams atom of graphite in air is :

\[\begin {array} {1 1} (1)\;3.01 \times 10^{22} & \quad (2)\;6.02 \times 10^{23} \\ (3)\;6.02 \times 10^{22} & \quad (4)\;3.01 \times 10^{23} \end {array}\]

1 Answer

$(3)\;6.02 \times 10^{22}$
answered Nov 7, 2013 by pady_1
 

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