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Sagot :
Let's use PO (resp PN) to denote the percentage of oxygen (resp nitrogen) and M for the molar mass.
A : PO/PN~1/2 -> We try N(2)O, which has M=2*14+16=44 which matches with the experimental data.
B : PO/PN~2 -> We try NO(2), which has M=14+2*16=46 which matches
C : PO/PN~3/2 -> We try N(2)O(3) which has M=14*2+3*16 76 which matches
D : PO/PN~5/2 -> We try N(2)O(3) which has M=14*2+5*16=108 which matches
E : PO/PN~1 -> We try NO which has M=30 which matches
A : PO/PN~1/2 -> We try N(2)O, which has M=2*14+16=44 which matches with the experimental data.
B : PO/PN~2 -> We try NO(2), which has M=14+2*16=46 which matches
C : PO/PN~3/2 -> We try N(2)O(3) which has M=14*2+3*16 76 which matches
D : PO/PN~5/2 -> We try N(2)O(3) which has M=14*2+5*16=108 which matches
E : PO/PN~1 -> We try NO which has M=30 which matches
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