A) 4.0 g
B) 1.0 g
C) 0.5 g
D) 2.0 g
Correct Answer: C
Solution :
: 32 g of\[{{O}_{2}}\]occupies 22.4 L at STP 4.0 g of\[{{O}_{2}}\]occupies\[\frac{22.4}{32}\times 4\]at STP = 2.8 L at STP 22.4 L volume is occupied by 4 g He 2.8 L volume is occupied by\[\frac{4}{22.4}\times 2.8\] = 0.5 got He.You need to login to perform this action.
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