calculate the volume of 4g of h2 at 4 atm
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Answered by
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The temperature is not given. I assume the standard temperature 273 K.
2 g of H2 at stp occupy 22.4 litres
4 g of H2 at stp occupy 44.8 litres.
P1V1 = P2V2
1 x 44.8 = 4 x V2 or V2 = 44.8/4 = 11.2 litres.
2 g of H2 at stp occupy 22.4 litres
4 g of H2 at stp occupy 44.8 litres.
P1V1 = P2V2
1 x 44.8 = 4 x V2 or V2 = 44.8/4 = 11.2 litres.
Answered by
1
(Assuming the Temperature to be 273K)
As we know PV = nRT
Thus V = nRT/P
n = no of moles = 4/2 =2 (molar mass of H2 =2)
Thus V = 2 x 0.082 x 273 / 4
= 11.193 Litres
Hope it helps.... Please mark as Brainliest
As we know PV = nRT
Thus V = nRT/P
n = no of moles = 4/2 =2 (molar mass of H2 =2)
Thus V = 2 x 0.082 x 273 / 4
= 11.193 Litres
Hope it helps.... Please mark as Brainliest
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