β‘οΈ Bir masalaning uch yechimi:
β195
#1 - usul:
Ar va OΒ² Mr=19*2=38
x + y = 1
40x+32y=38
x=0.75
y=0.25
Elektr toki Γ²tkazilganda kislorod ozonga aylanadi.
3OΒ² ---> 2OΒ³
1.5x x
Hosil bΓ²lgan aralashma tarkibi;
Ar 0.75mol ( 30g )
O2 0.25-1.5x (8-48x g)
O3 x mol (48x g)
Mr=20*2=40
30+8=38/40=0.95 bu hosil bo'lgan aralashmaning umumiy hajmi:
0.75/0.95=0.7895*100=78.95% β
#2-usul
D(H2)19*2=38
40 6 3mol
\. /
38
/. \
32 2 1mol
Elektr toki oβtganda O2 ozonga aylanadi.
3x 2x
3O2 β 2O3
1-3x qolgan O2
2x hosil boβlgan O3
3 mol Ar
D(H2) = 20*2=40
3*40+2x*48+(1-3x)*32
------------------------------------------ = 40
3+2x+1-3x
160-40x=152+96x-96x
X=0.2
0.2*2=0.4 mol O3
1-0.2*3=0.4 mol O2
3 mol Ar
Jami : 3.8 mol
3/3.8=78.95%
#Eng sodda 3-usuli:
38g/mol...........0.75 mol
40g/mol.........x=0.7895mol
0.7895β’100%=78.95%ββTelegram|
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