Question

va rog am nevoie mare de ajutor​

Answer 1

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=$\frac{100*3}{82}$=3,65% H

y= $\frac{100*31}{82}$=37,8% P

z=$\frac{100*48}{82}$=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=$\frac{100*2}{82}$=2,43% H

y=39,02% P

z= $\frac{100*48}{82}$=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