Question

( 22400 mathrm{ml} 00 mathrm{N}_{2}=28 mathrm{g} mathrm{sin} )
[
56 mathrm{m} / mathrm{O}_{2} mathrm{N}_{2}=frac{28 times 56}{22400}=0.07 mathrm{g}
]
Here weight of oxide ( (x)=0.11 g ) A ther reduction the weight ( =0.07 ) g ( log 5 )
sisnilarly weigut of oxides ( u(1 ; 0.15 mathrm{g} )
[
text { After reduction = 0 } cdot 07 mathrm{g}
]
( 0.07900 mathrm{N}_{2} ) peacts win
[
0.04 mathrm{g} 050_{2}
]
Hence, wit of ( N_{2} ) which would combine widn ( 0.08 mathrm{g} ) If 02 should be equal to
[
Rightarrow frac{0 cdot 0,7 times 0 cdot 08}{0.04}=0 cdot 14 g
]
In both cases (1) and
(2) the weight ( U B ) ( N_{2} ) which coonsines noith fixed amount ( sqrt{6} 02(0.049) operatorname{arco.07} g operatorname{ard} theta cdot 14 g )
Which are insimplest detio 2: 1 . Hence, the oride follows the law of multiple proportions.

# типрс оролцоно. 7. If a certain oxide of nitrogen weighing 0-11 g gives 56 mL of nitrogen and another oxide of nitrogen weighing 0.15 g gives the same volume of nitrogen (both at STP), show that these results support the law of multiple proportions. 10 -12L Db G .775 of red land Mh n yielded on otrona

Solution