A) 8 g
B) 16 g
C) 24 g
D) 32 g
Correct Answer: B
Solution :
\[pV=nRT=\frac{w}{m}Rt\] \[w=\frac{pVm}{RT}=\frac{12.315\times 1\times 32}{0.0821\times 300}=16g\] \[{{O}_{2}}\]to be released\[=32-16=16g\]You need to login to perform this action.
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