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MD20=20 g
1mol D20 =22.4dm3
40g D20 -- 44.8dm3 D2
400gD2O -- xdm3 D2
x=448dm3
odp. Powstanie 448dm3 deuteru
400gciężkiej wody-X
X= 448dm sześciennych deuteru
40g........44.8dm^3
2D20 ---> 2D2 + 02
400g.......x
x=400gx44.8dm^3/40g=448dm^3