a)Masa molara H3PO3=3*1+31+3*16=82g/mol
Dupa regula de 3 simpla vom spune ca
82 g H3PO3...3 g H...31 g P... 48 g O
100 g de H3PO3...x ...y ... z
x=[tex]\frac{100*3}{82}[/tex]=3,65% H
y= [tex]\frac{100*31}{82}[/tex]=37,8% P
z=[tex]\frac{100*48}{82}[/tex]=58,53%O
b) M H2SO3=2*1+32+3*16=82 g/mol
82 g H2SO3...2 g H...32 g S...48 g O
100 g H2SO3...x ... y ... z
x=[tex]\frac{100*2}{82}[/tex]=2,43% H
y=39,02% P
z= [tex]\frac{100*48}{82}[/tex]=58,53% O
c) M HNO3= 1+14+3*16=63 g/mol
63 g HNO3... 1 g H... 14 g N... 48 g 0
100 g HNO3... x ... y ... z
x=1,58% H
y= 22,22% N
z=76,19 % 0
d) M H2S= 2*1+32=34 g/mol
34 g H2S...2 g H... 32 g S
100 g H2S...x ... y
x=5,88% H
y= 94,11% S
e) M CO= 12+16=28 g/mol
28 g CO... 12 g C ... 16 g O
100 g CO... x ... y
x= 42,85 % C
y= 57, 14 % O
f) M MgO=24+16=40 g/mol
40 g MgO... 24 g Mg ... 16 g 0
100 g MgO... x .... y
x= 6o% Mg
y= 40 % O
iti sugerez sa mai faci odata calculele pentru regula de 3 simpla ca sa fii sigur ca sunt corecte
