`Equal masses of NO2 and O2 react completely leaving no reactants. Mole fraction of NO in the mixture containing only and NO and NO2 is X/8. find x ?(N2O O2 ->NO + NO2)`
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Answer:
2N
2
O
5
(g)→4NO
2
(g)+O
2
(g).
t=0 3.0M
t=30 2.75M
Now,
Δt
−Δ[N
2
O
5
]
=−
30−0
2.75−3
=
30
0.25
M/min
Again,
2
1
×
Δt
−Δ[N
2
O
5
]
=
4
1
×
Δt
−Δ[NO
2
]
Δt
−Δ[NO
2
]
=
30
0.25
×2=1.66×10
−2
M/min
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