calculate the volume of 6×10^23 H2 molecules
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Answered by
1
Answer:
22.4 litre
Explanation:
By STP we mean
P= 1 bar
T= 273.15 K
For an ideal gas
PV = nRT
R= 8.314 Lbar K⁻¹ mol⁻¹
6.023×10^{23}10
23
molecules makes 1 mole
n = 1
V = 22.4 L.
NOTE: 1 MOLE OF ANY GAS AT STP OCCUPIES 22.4 L.
Answered by
0
Answer:
22.4 litres
Explanation:
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