Question

please answer this question​

Answer 1

Explanation:

Rate of formation of n2 is same as n2o5, but rate of formation of O2 is 5/2*6.25*10–³

Answer 2

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2N2O5 -----> 2N2 + 5O2

-(1/2)∆(N2O5)/∆t = (1/2) ∆(N2)/∆t

= (1/5) ∆(O2)/∆t

Where ∆ repsent change with respect to time.

∆(N2O5)/∆t = 6.25 × 10-³mol L-¹ S-¹ ...{ Given }

∆(N2)/∆t = ?

And

∆(O2)/∆t = ?

=>

(1/2) × ( 6.25 × 10-³ ) = (1/2) × ∆(N2)/∆t

∆(N2)/∆t = 6.25 × 10-³ Mol L-¹ S-¹

And

(1/2)×(6.25 ×10-³ ) = (1/5) × ∆(O2)/∆t

∆(O2)/∆t = 15.625×10-³ Mol L-¹ S-¹