4NH3+502->4NO + 6H20 n=34/17=2
stosuje clapeyrona :
pv=nRT
V=nRT/p=2x83,1x286/1000=47,53 dm3
4 Nh3+ 5O2= 4NO+ 6h2O
n= 34/:17
n= 2
ze wzoru:
pv= nRT
V= nRT / :p
V= 2*83,1 * 286/:100
V= 47,5dm3
objętość tlenku azotu 2 wynosi 47,5 dm3
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4NH3+502->4NO + 6H20 n=34/17=2
stosuje clapeyrona :
pv=nRT
V=nRT/p=2x83,1x286/1000=47,53 dm3
4 Nh3+ 5O2= 4NO+ 6h2O
n= 34/:17
n= 2
ze wzoru:
pv= nRT
V= nRT / :p
V= 2*83,1 * 286/:100
V= 47,5dm3
objętość tlenku azotu 2 wynosi 47,5 dm3