Oblicz mase w warunkach normalnych:
a)12 dm3 amoniaku
b)600cm3 para bromu cząsteczkowego
c)2,5m3 wodoru
d)25,223dm3 argonu
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Oblicz mase w warunkach normalnych:
a)12 dm3 amoniaku
mNH3=17u
17g NH3-------22,4dm3
xg NH3---------12dm3
x = 9,11g NH3
b)600cm3 para bromu cząsteczkowego
mBr2=160u
160g Br2-----22,4dm3
xg Br2--------0,6dm3
x = 4,28g Br2
c)2,5m3 wodoru
mH2=2u
2g H2-------22,4dm3
xg H2-------2500dm3
x = 223,2g H2
d)25,223dm3 argonu
mAr=40u
40g Ar------22,4dm3
xg Ar-------25,223dm3
x = 45,04g Ar