Volume of 2.8 gm of nitrogen at STP in cc
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1 Mole of any gas at STP has a volume = 22.4L
This means 1 mole of N2 = 22.4L
28g of NE =22.4L. 【1 mole NE=28g】
So ,
1g=22.4÷28 L
2.8g = (22.4÷28)×2.8
=2.24 L
The retire 2.8g of NE at STP =2.24L
=2240 cc
❤Hope it helps mate |
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